Answer:
[tex]\large \boxed{\text{(a) 42 000 yr; (b) 45 000 yr}}[/tex]
Explanation:
Two important equations in radioactive decay are
[tex]\ln \dfrac{N_{0} }{N_{t}} = kt\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]
We use them for carbon dating.
(a) Initial activity = 0.23 Bq
(i) Calculate the rate constant
The half-life of ¹⁴C is 5730 yr.
[tex]\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 \times 10^{-4}\text{ yr}^{-1}\\\end{array}[/tex]
(ii) Calculate the age of the sample
[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.23 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 153 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.03 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.03}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{42 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{42 000 yr}}$}[/tex]
(b) Initial activity = 45 % larger
N₀ = 1.45 × 0.230 Bq = 0.334 Bq
[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.334 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 222 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.40 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.40}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{45 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{45 000 yr}}$}[/tex]
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by)2545(2121xxxxVwhere Vis in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and2VEin a mixture containing 40 mole percent of
The question is incomplete. The complete question is:
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
ΔV = x1x2(45x1 + 25x2)
Where ΔV is in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.
Answer:
1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹
Explanation:
In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:
ΔV = 0.4*0.6(45*0.4+ 25*0.6)
ΔV = 7.92 cm³.mol⁻¹
1VE - V1 = 7.92
1VE = 7.92 + 110
1VE = 117.92 cm³.mol⁻¹
2VE - V2 = 7.92
2VE = 7.92 + 90
2VE = 97.92 cm³.mol⁻¹
To determine the partial molar volumes of species 1 and 2 in a mixture containing 40 mole percent of species 1, we can use the equation: ΔV = ( V1E * x1) + ( V2E * x2). Plugging in the given values, the partial molar volumes of species 1 and 2 in the mixture are 94 cm3-mol-1 each.
Explanation:To determine the partial molar volumes of species 1 and 2 in a mixture containing 40 mole percent of species 1, we can use the equation:
ΔV = (V1E * x1) + (V2E * x2)
Where ΔV is the volume change of mixing, V1E and V2E are the partial molar volumes of species 1 and 2, and x1 and x2 are the mole fractions of species 1 and 2 respectively.
In this case, x1 = 0.40 and x2 = 0.60 (since the mixture contains 40-mole percent of species 1), V1E = 110 cm3-mol-1 and V2E = 90 cm3-mol-1. Plugging these values into the equation, we get:
ΔV = (110 cm3-mol-1 * 0.40) + (90 cm3-mol-1 * 0.60)
Simplifying this, we find that ΔV = 94 cm3-mol-1. Therefore, the partial molar volumes of species 1 and 2 in the mixture are 94 cm3-mol-1 and 94 cm3-mol-1 respectively.
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A 14 carbon fatty acid that goes through the fatty acid spiral will net how much energy, measured in ATP?
94 ATP molecules
Explanation:To answer the question we need to understand the beta oxidation of fatty acid.
First we need to know each electron carrier produces different number of ATP molecules in the electron transport chain.That is, FADH₂ yields 1.5 ATP while NADH yields 2.5 ATP moleculesAdditionally, for every acetyl coA, 10 ATP molecules are produced.What happens during oxidation of a fatty acid with even number of carbon, C₂ₙ?
During beta oxidation of an even number fatty acid, C₂ₙ, n-1 FADH₂, n-1 NADH and n acetyl coA molecules are produced.Therefore, for a fatty acid with 14 carbons, then 6 molecules of FADH₂, 6 molecules of NADH and 7 acetyl coA are produced.Therefore, the number of ATP molecules will be;From; 6 FADH₂ = 6 × 1.5 = 9 ATP
6 NADH; 6 × 2.5 = 15 ATP
7 Acetly CoA ; 7 × 10 = 70 ATP
This gives a total of 94 ATP molecules
Hence a 14 carbon fatty acid produces 94 ATP molecules after beta oxidation.
A 14 Carbon fatty acid has 94 ATP molecules.
Before answering this question we need to learn the ATP yield for every oxidation cycle which is:
FADH₂ produces 1.5 ATP while NADH yields 2.5 ATP molecules and for every acetyl CoA, 10 ATP molecules are produced.Beta Oxidation of Fatty Acid:
In beta-oxidation fatty acid molecules breaks down via catabolic process. During β-oxidation for even numbered fatty acid one molecule of FADH₂, NADH and acetyl CoA are formed.Fatty acid containing 14 carbon atoms produces 6 molecules of FADH₂, 6 molecules of NADH and 7 molecules of acetyl CoA.So for the calculation of ATP molecules:
6 FADH₂ =6×1.5=9 ATP
6 NADH=6×2.5=15 ATP
7 acetyl CoA=7×10=70 ATP
Total number of ATP molecules are 94.
Therefore, a 14 carbon fatty acid produces 94 ATP molecules.
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A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the sample was formed, how old is it? The half-life of U-238 is 4.5 × 109 years.1.4 × 1010 years2.1 × 109 years7.3 × 1011 years7.1 × 109 years2.4 × 1010 years
Answer:
2.1x10⁹ years
Explanation:
U-238 is a radioactive substance, which decays in radioactive particles. It means that this substance will lose mass, and will form another compound, the Pb-206.
The time need for a compound loses half of its mass is called half-life, and knowing the initial mass (mi) and the final mass (m) the number of half-lives passed (n) can be found by:
m = mi/2ⁿ
The mass of Pb-206 will be the mass that was lost by U-238, so it will be mi - m. Thus, the mass ration can be expressed as:
(mi-m)/m = 0.337/1
mi - m = 0.337m
mi = 1.337m
Substituing mi in the expression of half-life:
m = 1.337m/2ⁿ
2ⁿ = 1.337m/m
2ⁿ = 1.337
ln(2ⁿ) = ln(1.337)
n*ln(2) = ln(1.337)
n = ln(1.337)/ln2
n = 0.4190
The time passed (t), or the age of the sample, is the half-life time multiplied by n:
t = 4.5x10⁹ * 0.4190
t = 1.88x10⁹ ≅ 2.1x10⁹ years
The age of a geological sample, given a Pb-206/U-238 mass ratio, can be calculated using the principles of radioactive decay and the known half-life of U-238. The formula t = (1/λ) × ln(1 + (Pb-206/U-238)) is used, where λ is the decay constant calculated as 0.693 / half-life.
Explanation:To calculate the age of a geological sample based on the Pb-206/U-238 mass ratio and the half-life of U-238, we need to apply the principles of radioactive decay.
U-238 decays into Pb-206 with a half-life of 4.5 × 10⁹ years. The decay constant (λ) can be calculated, as λ = 0.693 / half-life. Assuming that there was no Pb-206 present when the sample was formed, we can derive the time passed since the rock formed using the formula t = (1/λ) × ln(1 + (Pb-206/U-238)).
Given that the Pb-206/U-238 mass ratio is 0.337/1.00, we would insert these values into the above formula to calculate the age. Without actual numeric calculation of this equation, we cannot provide a specific number among the options listed, but this is the method you would use to do so.
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Why are electrolytes important in your body? Do you need to drink Gatorade to get them?
Electrolytes are substances -minerals- that conduct electricity once they are dissolved in water. Some examples include Na⁺ (Sodium), Ca⁺² (Calcium), PO₄⁻³ (Phosphate) and Mg⁺² (Magnesium).
They are important because of the rols they play in our organism: The electrical charges they carry are what allow muscles to contract and nerve impulses to be transmitted. Without proper muscle contractions or nerve transmissions we would die.
We do not need to drink Gatorade to get electrolytes. It's true that sport drinks contain electrolytes, but those electrolytes can also come from different sources in our everyday diet, such as salt or fruits and vegetables.
A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [ Na + ] = M [ NO − 3 ] = M [ Sr 2 + ] = M [ F − ] = M
Answer:
Mass SrF2 produced = 38.63 g SrF2 produced
[Na^+]: = 1.758 M
[NO3^-]: = 1.238 M
[Sr^2+] = 0.3589 M
[F^-] = 2.36*10^-5 M
Explanation:
Step 1: Data given
Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L
Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L
Step 2 : The balanced equation
Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2
Step 3: Calculate moles strontium nitrate
Moles Sr(NO3)2 = Molarity * volume
Moles Sr(NO3)2 = 2.888 M * 0.150 L
Moles Sr(NO3)2 = 0.4332 moles
Step 4: Calculate moles NaF
Moles NaF = 3.076 M * 0.200 L
Moles NaF = 0.6152 moles
It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.
Step 5: Calculate limiting reactant
For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3
NaF is the limiting reactant. It will completely be consumed (0.6152 moles).
Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles
Moles Sr^2+ precipitated by F^- = 0.3076
There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2
Moles Sr^2+ no precipitated (left over) = 0.1256 moles
Step 6: Calculate moles SrF2
For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3
For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2
Mass SrF2 produced: 0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced
Step 7: Calculate concentration of [Na+] and [NO3-]
Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:
[Na^+]: (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M
[NO3^-]: (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M
Step 8: Calculate [Sr^2+] and [F^-]
[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M
To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.
SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)
Ksp = [Sr^2+][F^-]²
2x10^-10 = (0.3589)(x)²
x² = 5.57*10^-10
x = [F^-] = 2.36*10^-5 M
The mass of SrF2 precipitate formed in the reaction is 38.69 g
We have to first write down the balanced reaction equation as follows;
Sr(NO3)2(aq) + 2NaF(aq)------> SrF2(s) + 2NaNO3(aq)
Next, we obtain the number of moles of each reactant;
Amount of Sr(NO3)2 = 150.0/1000 × 2.888 = 0.433 mols of Sr(NO3)2
Amount of NaF = 200.0/1000 × 3.076 = 0.615 moles of NaF
We have to obtain the limiting reactant. This is the reactant that yields the lowest number of moles of product.
For Sr(NO3)2:
1 mole of Sr(NO3)2 yields 1 mole of SrF2
0.433 mols of Sr(NO3)2 yields 0.433 mols of SrF2
For NaF;
2 moles of NaF yields 1 mole of SrF2
0.615 moles of NaF yields 0.615 × 1/2 = 0.308 moles of SrF2
Hence NaF is the limiting reactant.
Mass of SrF2 formed = 0.308 moles of SrF2 × 125.62 g/mol = 38.69 g
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Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are
shown in the balanced equation
Answer:
Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
Explanation:
Decomposition reaction:
When a single compound break down into two or more simpler products.
For example "AB" reactant undergoes decomposition to form "A" and "B" products.
The chemical reaction is as follows.
[tex]AB\rightarrow A+B[/tex]
The given compound is aluminium oxide.
The decomposition reaction of aluminium oxide is a follows.
[tex]Al_{2}O_{3}\rightarrow Al+O_{2}[/tex]
The balanced equation is as follows.
[tex]2Al_{2}O_{3}\rightarrow 4Al+3O_{2}[/tex]
Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If the heat released during condensation goes only toward heating the metal, what is the final temperature (in degrees Celsius) of the metal block? (The specific heat capacity of aluminum, C s,Al , is 0.903 J/(g⋅ ∘ C) .) Express the temperature in degrees Celsius to two significant figures.
Answer:
[tex]49^oC[/tex]
Explanation:
At [tex]25^oC[/tex], the heat of vaporization of water is given by:
[tex]\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g[/tex]
The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:
[tex]Q_1 = \Delta H^o_{vap} m_w[/tex]
The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:
[tex]Q_2 = c_{Al}m_{Al}(t_f - t_i)[/tex]
According to the law of energy conservation, the heat lost is equal to the heat gained:
[tex]Q_1 = Q_2[/tex] or:
[tex]\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)[/tex]
Rearrange for the final temperature:
[tex]\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i[/tex]
We obtain:
[tex]\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f[/tex]
Then:
[tex]t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC[/tex]
To find the final temperature of the aluminum block, we can use the equation for heat transfer: Q = mcΔT. By substituting the mass of the water, specific heat capacity of aluminum, and initial temperature into the equation, we can calculate the temperature change (ΔT) of the aluminum block and then add it to the initial temperature to find the final temperature.
Explanation:To find the final temperature of the aluminum block, we can use the equation for heat transfer: Q = mcΔT. The heat lost by the aluminum block is equal to the heat released during condensation of the water. We know the mass of the water (0.48 g) and the specific heat capacity of aluminum (0.903 J/(g⋅°C)). By substituting these values into the equation, we can calculate the temperature change (ΔT) of the aluminum block and then add it to the initial temperature (25°C) to find the final temperature.
First, we calculate the heat lost by the aluminum block: Q = (mass of water) × (specific heat capacity of aluminum) × (final temperature - initial temperature). Q = 0.48 g × 0.903 J/(g⋅°C) × ΔT.
Since the heat released during condensation only goes toward heating the metal, the heat lost by the aluminum block is equal to the heat gained by the water: Q = mwater × Cwater × ΔT. By substituting the known values and solving for ΔT, we can then find the final temperature by adding ΔT to the initial temperature of the aluminum block.
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An unknown compound with empirical formula C3H5 was treated with Br2/CCl4. The bromine solution went from orangish/red to clear immediately at room temperature. Upon treatment with O3 followed by work-up with dimethylsulfide the following products were identified. From the information provided what is/are the most likely structure(s) for this unknown compound.
Answer:
(E)-3-methylpenta-1,3-diene
Explanation:
Bromine water in carbon tetrachloride is used to test for the presence of alkenes. Bromine was added to this unsaturated compound with double bonds, as the reddish color of the solution disappeared. This means bromine was added to the double bonds present in our structure.
Since three products were formed upon ozonolysis, this means we have two double bonds present in our structure, a diene. Besides, a total of 6 carbons are seen in the given structures, so molecular formula of the compound is [tex]C_6H_{10}[/tex].
Ozonolysis occurs when a double bond is broken and oxygen atoms are added to each end of the double bond, that's why we have a total of 3 products.
Our starting material is (E)-3-methylpenta-1,3-diene. You may wish to refer to the file uploaded below which represents the ozonolysis step of breaking the two double bonds and forming the 3 products shown.
Final answer:
The unknown compound undergoing the given reactions is likely to have the structures (1'R,2'S,3'R)-1-[(2',3'-O-isopropylidene)-4'-cyclopenten-1'-yl]-4-iodoimidazole and 06-(Benzotriazol-1-yl)-2-bromo-9-[2-deoxy-3,5-di-O-(t-butyldimethylsilyl)-B-D-ribofuranosyllpurine.
Explanation:
The unknown compound with empirical formula C3H5 reacted with Br2/CCl4 to undergo an addition reaction. This is indicated by the change in color of the bromine solution from orangish/red to clear. Further treatment with O3 followed by work-up with dimethylsulfide resulted in the identification of specific products. Based on this information, the most likely structures for the unknown compound are: (1'R,2'S,3'R)-1-[(2',3'-O-isopropylidene)-4'-cyclopenten-1'-yl]-4-iodoimidazole and 06-(Benzotriazol-1-yl)-2-bromo-9-[2-deoxy-3,5-di-O-(t-butyldimethylsilyl)-B-D-ribofuranosyllpurine.
Uranium-234 is unstable and undergoes two subsequent alpha emissions. What is the resulting nuclide from this transformation?
Answer:Radium
Explanation:
The nuclear reaction involving two alpha emissions of 234 U is shown in the diagram. This leads to the formation of a 226Ra nucleus.
After two alpha emissions, uranium-234 transforms into radium-226, as each alpha emission reduces the atomic number by 2 and the mass number by 4.
Explanation:Uranium-234 (U-234) undergoes two consecutive alpha emissions, which means it emits two alpha particles. An alpha particle is identical to a helium nucleus and is composed of 2 protons and 2 neutrons. So, each alpha emission will reduce the atomic number of the parent nuclide by 2 and the mass number by 4.
Since U-234 has 92 protons and 234 nucleons, after two alpha emissions it will have 88 protons and 226 nucleons. The resulting element with 88 protons is radium (Ra), so the resulting nuclide after two alpha decays of uranium-234 is radium-226 (Ra-226).
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Compete the following table comparing atomic mass and mass number by enteng Yes or No in each box.
Answer:
atomic mass; mass number
1) NO ; YES
2) YES ; YES
3) YES ; YES
4) YES ; NO
5) NO; YES
Explanation:
An atomic mass is the mass of a single atom of a chemical element. It includes the masses of the 3 subatomic particles that make up an atom: protons, neutrons and electrons.
And also 1 atomic mass unit is defined as [tex]\frac{1}{12}[/tex] of the mass of a single carbon-12 atom.
Now, mass number of an element is the sum of protons and neutrons present in a single atom of that element.
Mass of the electrons is 9.10938356 × 10⁻³¹ kilograms
which is negligible.
And mass of proton and neutron is nearly but not exactly 1 u.
so, mass number is nearly but not exactly equal to atomic mass ,most of the times.
so,
1) only mass number can be used to calculate number neutrons present by simple subtracting atomic number(proton number) in mass number
2) both of these are found in periodic table
3) both can be found for individual atoms
4) mass number are different for isotopes but the atomic mass is calculated considering the isotopes(depends on availability of isotopes)
5)only mass number is given in isotopic symbol.
If you started with 0.183 mol of N2 and 0.549 mol of H2, and they completely reacted in the reaction vessel, determine the total moles of gas particles (n) there are during the initial and final conditions. Additionally, determine the ratio of the number of gas particles in the products to that of the reactants, then complete the statements below.
Answer:
Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).
In the products we have 0.366 moles of gas
Explanation:
Step 1: Data given
Number of moles of N2 = 0.183 mol
Number of moles of H2 = 0.549 mol
Step 2: The balanced equation:
N2 + 3H2 → 2NH3
Step 3: Calculate the limiting reactant
For 1 mol of N2 we need 3 moles of H2 to produce 2 moles of NH3
There is no limiting reactant. But will be completely consumed
Step 4: Calculate moles of NH3
For 1 mol of N2 we have 2 moles of NH3 produced
For 0.183 moles of N2 we have 2*0.183 = 0.366 moles of NH3 produced
Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).
In the products we have 0.366 moles of gas
Answer:
Initially, there are 0.732 mol of gas particles. After the reaction is complete, there are 0.366 mol of gas particles. Therefore, the ratio of product to reactant prticles is 5.0
Explanation:
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the thermometer reaches steady state, it is suddenly placed in a bath at 110oF at t = 0 and left there for 1 min, after which it is immediately returned to the bath at 100oF.(a) Plot the variation of the thermometer reading with time (No hand-sketched plot please).(b) Calculate the thermometer reading at t = 0.5 min and at t = 2.0 min.
Answer:
(a) See below
(b) 103.935 °F; 102.235 °F
Explanation:
The equation relating the temperature to time is
[tex]T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )[/tex]
1. Calculate the thermometer readings after 0.5 min and 1 min
(a) After 0.5 min
[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}[/tex]
(b) After 1 min
[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}[/tex]
2. Calculate the thermometer reading after 2.0 min
T₀ =106.321 °F
ΔT = 100 - 106.321 °F = -6.321 °F
t = t - 1, because the cooling starts 1 min late
[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}[/tex]
3. Plot the temperature readings as a function of time.
The graphs are shown below.
To plot the variation of the thermometer reading with time, we need to consider the first-order dynamics of the thermometer. The time constant of the thermometer is 1 min, and it takes approximately 5 time constants to reach steady state. To calculate the thermometer reading at specific times, we can use the formula for first-order dynamics.
Explanation:To plot the variation of the thermometer reading with time, we need to consider the first-order dynamics of the thermometer. The time constant of the thermometer is 1 min, so it takes approximately 5 time constants (or 5 minutes) to reach steady state. After that, when the thermometer is suddenly placed in a bath at 110oF, it will start to change its reading towards the new temperature following the first-order dynamics. Then, at t = 1 min, it is returned to the bath at 100oF and again it starts to change its reading towards the new temperature.
(b) To calculate the thermometer reading at t = 0.5 min, we can use the formula for first-order dynamics: R = R0 + (R1 - R0)(1 - e^(-t/tau)), where R0 is the initial reading, R1 is the final reading, t is the time, and tau is the time constant. Plugging in the values, we get R = 100 + (110 - 100)(1 - e^(-0.5/1)) ≈ 105.3oF. To calculate the thermometer reading at t = 2.0 min, we use the same formula with t = 2.0 and the new R0 would be the reading at t = 1 min, which we can calculate similarly.
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The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the environment. Therefore, two containers of waste solution are separated by a semipermeable membrane and pressure is applied to one container, forcing only water molecules through the membrane and diluting the waste solution in the other container. As dilution continues, higher and higher pressures are needed to counteract the natural tendency for the water molecules to have a net flow back toward the more concentrated solution. What was the applied pressure at the end of this process if the final concentrations of the solutions were 0.053 M and 0.170 M at a temperature of 23 ∘C?
The applied pressure at the end of this process is P = -2.859atm
Explanation:
Write down the given values
there are concentration of two solutions
S1 = 0.053 M
S2 = 0.170 M
Temperature (T) = 23°C
that is written as 298 K
it can be solved by multiplying ans subtracting the values
P=MR1-MR2
0.053 M * 0.082 * 298 - 0.170 M * 0.082 * 298
= 1.295 - 4.15412
P = -2.859atm
Final answer:
Reverse osmosis is a technique that involves applying pressure to force water molecules from a high concentration solution to a low concentration solution through a semipermeable membrane. This method is commonly used to convert saltwater into freshwater. The concentration of the solutions will determine the amount of pressure required.
Explanation:
When a compartment containing a dilute solution is connected to another compartment containing a concentrated solution by a semipermeable membrane, water molecules move from the dilute solution to concentrated solution. This phenomenon is called osmosis. By applying pressure in the higher concentration solution, water molecules migrate from a high concentration solution to a low concentration solution through a semipermeable membrane. This method is called reverse osmosis water filter system. In this technique, the membrane must be able to tolerate the high pressure, and prevent solute molecules to pass through. This technology certainly works, and it has been used to convert salt (ocean or sea) water into fresh water. With this technique, the water with higher concentration is discharged. Thus, this technology is costly in regions where the water cost is high.
Which substance in each of the following pairs would you expect to have the higher boiling point? Explain why.
a. Ne or Xe
b. CO2 or CS2
c. CH4 or Cl2
d. F2 or LiF
e. NH3 or PH3
Explanation:
Boiling point is defined as the temperature when the vapor pressure of liquid becomes equal to the atmospheric pressure surrounding the liquid. And, during this temperature liquid state of substance changes into vapor state.
More stronger is the presence of intermolecular forces within the molecules of a compound more heat is required by it to break the bond. Hence, boiling point will also increase.
(a) As both Ne and Xe are noble gases and has intermolecular dispersion forces. But as the atomic size of Xe is more than the atomic size of Ne and we know that dispersion forces increases with increase in size.
Hence, boiling point of Xe will be more than the boiling point of Ne.
(b) Both [tex]CO_{2}[/tex] and [tex]CS_{2}[/tex] are non-polar in nature with intermolecular dispersion forces. So, more is the molecular weight of a compound more heat will be required by it in order to break the bonds.
As molecular weight of [tex]CS_{2}[/tex] is more than the molecular weight of [tex]CO_{2}[/tex]. Hence, [tex]CS_{2}[/tex] will have high boiling point.
(c) Molecular weight of [tex]CH_{4}[/tex] is 16 g/mol and molecular weight of [tex]Cl_{2}[/tex] is 35 g/mol.
Hence, [tex]Cl_{2}[/tex] will have high boiling point due to the presence of high molecular weight as compared to [tex]CH_{4}[/tex] and more strong dispersion forces.
(d) [tex]F_{2}[/tex] is a covalent compound so, it will also be non-polar in nature. Hence, it has weak intermolecular dispersion forces. On the other hand, LiF is an ionic compound and it has strong intermolecular forces of attraction due to the presence of opposite charges on the combining atoms.
Hence, LiF will have high boiling point as compared to [tex]F_{2}[/tex].
(e) Both [tex]NH_{3}[/tex] and [tex]PH_{3}[/tex] are polar molecules in which dipole-dipole forces does exist. Since, nitrogen atom is more electronegative than phosphorous atom so, it will have stronger hydrogen bonding and strong intermolecular forces.
Hence, [tex]NH_{3}[/tex] will have high boiling point as compared to [tex]PH_{3}[/tex].
The boiling point of Xe will be greater than boiling point of Ne.
Boiling point:
It is the temperature a which the vapor pressure of liquid is equal to the atmospheric pressure, surrounding the liquid. During this temperature, liquid start to vaporize.
The Boiling point depends upon the strength of intermolecular bond.
Since, Ne and Xe are noble gases and the atomic size of Xe is greater than Ne. hence they has intermolecular dispersion force of Ne will be greater than that of Xe.
Therefore, the boiling point of Xe will be more than that of Ne.
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Determine the identity of the daughter nuclide from the beta decay of 8938Sr.9039Y8536Kr8734Se9038Sr8939Y
Answer: The daughter nuclide formed by the beta decay of given isotope is [tex]_{39}^{89}\textrm{Y}[/tex]
Explanation:
Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.
The released beta particle is also known as electron.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
We are given:
Parent isotope = [tex]_{38}^{89}\textrm{Sr}[/tex]
The chemical equation for the beta decay process of [tex]_{38}^{89}\textrm{Sr}[/tex] follows:
[tex]_{38}^{89}\textrm{Sr}\rightarrow _{39}^{89}\textrm{Y}+_{-1}^0\beta[/tex]
Hence, the daughter nuclide formed by the beta decay of given isotope is [tex]_{39}^{89}\textrm{Y}[/tex]
Final answer:
The daughter nuclide from the beta decay of Strontium-89 is Yttrium-89, and the nuclear equation for the decay is 8938Sr → 8939Y + β-.
Explanation:
The question is asking to identify the daughter nuclide from the beta decay of Strontium-89 (8938Sr). In beta decay, a neutron in the nucleus is converted into a proton and a beta particle (an electron) is emitted. The atomic number increases by 1, but the mass number remains the same. Therefore, the daughter nuclide of 8938Sr undergoing beta decay would be 8939Y (Yttrium-89).
The nuclear equation for this decay process would be written as:
8938Sr → 8939Y + β-
Consider the following general voltaic cell and a cell notation, and answer all three parts of this question.
Mg(s) | Mg2+(aq) || Cl2(g) | Cl−(aq) | C(s)
Part 1: Based on the cell notation, the substance of Electrode A is [ Select ] ["Mg(s)", "C(s)", ""] , and the substance of Electrode B is [ Select ] ["C(s)", "Mg(s)"] .
Part 2: What is the balanced redox equation for the voltaic cell?
A) Mg2+(aq) + 2 Cl−(aq) \rightarrow → Mg(s) + Cl2(g)
B) Mg2+(aq) + Cl−(aq) \rightarrow → Mg(s) + Cl2(g)
C) Mg(s) + Cl2(g) \rightarrow → Mg2+(aq) + 2 Cl−(aq)
D) Mg(s) + Cl2(g) \rightarrow → Mg2+(aq) + Cl−(aq)
E) None of the above is correct, because C(s) doesn't appear in any of them.
Part 3:
- If the salt bridge contains NaNO3, Na+ ions flow to the [ Select ] ["left", "right"] , and NO3− ions flow to the [ Select ] ["left", "right"] .
- Electrons flow from Electrode [ Select ] ["A", "B"] to Electrode [ Select ] ["B", "A"] .
Answer:
Part -1
Electrode A - Mg(s)
Electrode B - [tex]Cl_{2}[/tex]
Part-2
Option -C
Part - 3
Electrons flow from electrode -A to electrode -B.
Explanation:
Part-1
Electrode A is [tex]Mg(s)|Mg^{+2}(aq)[/tex]
Electrode B is [tex]Cl_{2}(g)|Cl^{-}(aq)|C(s)[/tex]
Part-2
From the electrode, magnesium looses two electrons and chlorine gains two electrons. Therefore, balanced redox equation for the voltaic cell is as follows.
[tex]Mg(s)+Cl_{2}(g)\rightarrow Mg^{2+}(aq)+2Cl^{-}(aq)[/tex]
Part-3
[tex]Na^{+}[/tex] ions flow from the left to right.
[tex]NO_{3}^{-}[/tex] ions flow from the right to left.
Therefore, Electrons flow from electrode -A to electrode -B.
In the given voltaic cell notation, Electrode A is Magnesium (Mg) and Electrode B is Carbon (C). The correct balanced redox equation is Mg(s) + Cl₂(g)
ightarrow → Mg2+(aq) + 2 Cl−(aq). If the salt bridge contains NaNO3, Na+ ions flow to the right, and NO3− ions flow to the left, to maintain charge balance. Electrons flow from Electrode A to Electrode B.
Part 1: In the given cell notation, the substance of Electrode A is Mg(s), and the substance of Electrode B is C(s). Cell notation is written from left to right, starting with the anode (the electrode where oxidation occurs) and ending with the cathode (the electrode where reduction occurs). Thus, Magnesium (Mg) is Electrode A and Carbon (C) is Electrode B.
Part 2: The balanced redox equation for the given voltaic cell is Mg(s) + Cl₂(g)
right arrow → Mg₂ +(aq) + 2 Cl−(aq). This equation represents the overall process of the cell. Magnesium (Mg) is oxidized to Mg₂+ ions, releasing two electrons, and chlorine gas (Cl₂) is reduced to chloride ions (Cl-), accepting these electrons.
Part 3: If the salt bridge contains NaNO₃, Na+ ions flow to the right, and NO₃− ions flow to the left. The function of the salt bridge is to maintain charge balance because the reduction at the cathode makes the solution negative and the oxidation makes the anode side solution positive. In terms of electron flow, electrons flow from Electrode A to Electrode B.
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Indicate whether each statement is true or false. If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. Exothermic reactions are faster than endothermic reactions. If you double the temperature for a reaction, you cut the activation energy in half.
Answer:
False, False, False
Explanation:
Indicate whether each statement is true or false.
If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. FALSE. The relation between the rate constant (k) and the activation energy (Ea) is given by the Arrhenius equation, that is, k depends on the Ea, but not on the enthalpy (ΔH).Exothermic reactions are faster than endothermic reactions. FALSE. Whether a reaction is exothermic or endothermic is defined by the sign of the enthalpy. However, as stated previously, the rate constant (and the rate reaction) do not depend on the enthalpy of the reaction.If you double the temperature for a reaction, you cut the activation energy in half. FALSE. The activation energy does not depend on the temperature.When the rate constant should be determined so we cannot calculate the enthalpy change.
The Exothermic reactions should not be faster.
Indication of the statement:here the relationship that lies between the rate constant and the activation energy that should be provided by the Arrhenius equation should be based on the Ea.
In the case when the reaction should be considered as the exothermic so it represent the sign of the enthalpy.
Also, the activation energy does not based upon the temperature.
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For many purposes we can treat ammonia as an ideal gas at temperatures above its boiling point of . Suppose the pressure on a sample of ammonia gas at is tripled.
Is it possible to change the temperature of the ammonia at the same time such that the volume of the gas doesn't change?
A. yes
B. no
If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Answer:
A. yes
594°C
Explanation:
There is some info missing. I think this is the original question.
For many purposes we can treat ammonia (NH₃) as an ideal gas at temperatures above its boiling point of -33 °C. Suppose the pressure on a 6.0 m³ sample of ammonia gas at 16.0°C is tripled. Is it possible to change the temperature of the ammonia at the same time such that the volume of the gas doesn't change? If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Given data
V₁ = V₂ = 6.0 m³
T₁ = 16.0°C + 273.15 = 289.2 K
T₂ = ?
P₂ = 3 P₁
Assuming constant volume and ideal gas behavior, we can find the new temperature using the Gay-Lussac's law.
[tex]\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}} \\T_{2}=\frac{P_{2}.T_{1}}{P_{1}} =\frac{3P_{1}.T_{1}}{P_{1}}=3T_{1}=3 \times 289.2K = 867.6K[/tex]
°C = 867.6K - 273.15 = 594°C
Yes, it is possible to change the temperature of the ammonia gas while keeping the volume constant. The new temperature is approximately -88.6 °C.
Explanation:According to Gay-Lussac's law, for a given amount of gas at constant volume, the pressure and temperature are directly proportional. Therefore, if the pressure is tripled, the temperature of the ammonia gas can be changed in the same proportion to keep the volume constant.
We can use the formula P1/T1 = P2/T2 to calculate the new temperature (T2) when the pressure (P) is tripled. Given that the temperature (T1) is the boiling point of ammonia, we can plug in the values and solve for T2.
Using this formula, we can find that the new temperature is approximately -88.6 °C.
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The balanced reaction equation for combustion of heptane, C 7 H 16 , is C 7 H 16 + 11 O 2 ⟶ 7 CO 2 + 8 H 2 O If the reaction produced 23.5 g CO 2 , how many grams of heptane were burned? mass: g C 7 H 16 The reaction of limestone with hydrochloric acid is CaCO 3 + 2 HCl ⟶ CaCl 2 + CO 2 + H 2 O If the reaction produced 23.5 g CO 2 , how many grams of HCl reacted? mass: g HCl
Answer:
1) There were 7.65 grams of heptane burned
2) There reacted 38.94 grams of HCl
Explanation:
1) The combustion of heptane
Step 1: Data given
Mass of CO2 = 23.5 grams
Molar mass of CO2 = 44.01 g/mol
Step 2: The balanced equation:
C7H16 + 11O2 ⟶ 7CO2 + 8H2O
Step 3: Calculate moles of CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 23.5 grams / 44.01 g/mol
Moles CO2 = 0.534 moles
Step 4: Calculate moles heptane
For 1 mole of Heptane , we need 11 moles of O2 to produce 7 moles of CO2 and 8 moles of H2O
For 0.534 moles of CO2 we have 0.534/7 = 0.0763 moles of heptane
Step 5: Calculate mass of heptane
Mass of heptane = moles heptane * molar mass heptane
Mass heptane = 0.0763 moles * 100.21 g/mol
Mass heptane = 7.65 grams
2) The reaction of limestone with hydrochloric acid
Step 1: Data given
Mass of CO2 = 23.5 grams
Step 2: The balanced equation:
CaCO3 + 2HCl ⟶ CaCl2 + CO2 + H2O
Step 3: Calculate moles of CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 23.5 grams / 44.01 g/mol
Moles CO2 = 0.534 moles
Step 4: Calculate moles of HCl
For 1 mol of CaCO3 we need 2 moles of HCl to produce 1 mol of CaCl2, 1 mol of CO2 and 1 mol of H2O
For 0.534 moles of CO2 we have 2*0.534 = 1.068 moles of HCl
Step 5: Calculate mass of HCl
Mass HCl = moles HCl * molar mass HCl
Mass HCl = 1.068 moles * 36.46 g/mol
Mass HCl = 38.94 grams
To produce 23.5 g of CO₂, approximately 7.63 g of heptane were burned, and 38.94 g of HCl reacted. By using the molar masses and stoichiometric relationships from the balanced reactions, we can find the required masses. Moles of substances involved were calculated using the molar masses and balanced equations.
Determining Grams of Heptane and HCl Reacted
First, let's find out how many grams of heptane (C₇H₁₆) were burned to produce 23.5 g of CO₂ in the given balanced chemical reaction:
Combustion of Heptane:
Balanced equation: C₇H₁₆ + 11 O₂ ⟶ 7 CO₂ + 8 H₂O
Molar mass of CO₂: 12.01 (C) + 2 × 16.00 (O) = 44.01 g/molMoles of CO₂ produced: 23.5 g ÷ 44.01 g/mol ≈ 0.534 moles CO₂From the balanced equation, 7 moles of CO₂ are produced per 1 mole of heptane. Thus, moles of heptane burned: 0.534 moles CO₂ ÷ 7 ≈ 0.076 moles heptane.Molar mass of heptane (C₇H₁₆): 7 × 12.01 (C) + 16 × 1.01 (H) = 100.23 g/molGrams of heptane burned: 0.076 moles × 100.23 g/mol ≈ 7.63 g.Next, let's determine how many grams of HCl reacted to produce 23.5 g of CO₂ in the reaction between limestone and hydrochloric acid:
Reaction of Limestone with Hydrochloric Acid:
Balanced equation: CaCO₃ + 2 HCl ⟶ CaCl₂ + CO₂ + H₂O
Moles of CO₂ produced: 23.5 g ÷ 44.01 g/mol ≈ 0.534 moles CO₂From the balanced equation, 1 mole of CO₂ is produced per 2 moles of HCl. Thus, moles of HCl reacted: 0.534 moles CO₂ × 2 ≈ 1.068 moles HCl.Molar mass of HCl: 1.01 (H) + 35.45 (Cl) = 36.46 g/molGrams of HCl reacted: 1.068 moles × 36.46 g/mol ≈ 38.94 g.Sodium carbonate solution reacts with strontium chloride solution. Predict whether the reaction will produce a precipitate.
Answer:
SrCO₃ will precipitate in the reaction
Explanation:
Na₂CO₃ + SrCl₂ → 2 NaCl + SrCO₃ ↓
CO₃⁻² reacts with all the elements of group 2 of periodic table, to make an insoluble solid.
NaCl is a ionic salt, which is extremely soluble.
To solve this, you have to analyse all the reactions of cations and anions, with the qualitative inorganic analysis.
An ether solution of PhCO2H (A), PhNH2 (B), and PhCH3 (C) is extracted with aqueous NaOH. The ether layer will contain what compound(s) after the extraction?
Answer:
The ether layer will contain aniline and toluene after the extraction.
Explanation:
A solvent extraction is a method to separate compounds or metal complexes, based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent (Ether, in this case).
PhCO₂H (Benzoic acid) reacts with NaOH to form benzoate ion that is very soluble in water but poorly soluble in ether. That means PhCO₂H will be in the aqueous layer.
PhNH₂ (Aniline) is very soluble in organic solvents as ether but poorly soluble in water,
PhCH₃ (Toluene) is soluble in ether but insoluble in water.
That means the ether layer will contain aniline and toluene after the extraction.
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3. How was the rainbow created by the light emitting from the flashlight?
Answer:
by nuclear reaction
Explanation:
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The light refracts and is split into the spectrum colours.
In simple terms, the light splits into the colours.
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Apply the given electronegativities to each bond in the examples below to see which are polar and which are nonpolar. Multiple bonds are treated the same as single bonds.
H-O-H
O-C-O
H-C-N
Answer:
H-O-H polar
O-C-O nonpolar
H-C-N polar
Explanation:
Looking up the electronegativities of the atoms involved in this question, we have:
Atom Electronegativity
H 2.2
C 2.55
N 3.04
O 3.44
All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.
In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H of the ideal tetrahedral structure to be smaller than 109.5 º resulting in an overall dipole moment making it polar.
In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule since the central carbon atom does not have lone pairs of electrons since it has 2 double bonds. Therefore the molecule is nonpolar.
In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.
Answer:
H-O-H is polar
O-C-O is non polar
H-C-N is polar
Explanation:
H-O-HThe electronegativity value of H is 2.1 and the Electronegavity value for O is 3.5 which makes the electronegativity difference of 1.4. O has 2 lone pair of electrons which makes the shape non- linear and polar.
O-C-OThe bond between C and O is a double bond. The Electronegativity of O is 3.5 while the Electronegativity value for C is 2.5 which gives the electronegativity difference of 1.0. There are no lone pairs on carbon and the dipole or partial charges on carbon and oxygen cancels out making O-C-O non-polar.
H-C-NThe Electronegativity value of H is 2.1, electronegativity value for C is 2.5 and the electronegativity value for N is 3.0 which makes the molecule to develop partial positive charge on H and partial negative charge on the C-N making the compound polar.
Geometrically tetrahedral means that the electron groups have what angle? A. 120° B. 180° C. 90° D. 109.5°
Answer:
109.5° (d)
Explanation:
Use the table 10.1 to determine the
1. Electron geometry
2. Molecular geometry
3. Bond angles
-Four electron groups give a tetrahedral electron geometry
- two bonding groups and two lone pair give a bent molecular geometry
- the idealized bond angles for tetrahedral geometry are 109.5°.
The given potentials are observed for a calomel electrode: E ° = 0.268 V and E ( saturated KCl ) = 0.241 V . Use this information to predict the value of E for a silver-silver chloride electrode saturated with KCl . For the silver-silver chloride electrode, E ∘ = 0.222 V .
The potential of the silver-silver chloride electrode in a saturated KCl solution can be predicted as 0.195V, whatever the potential of the calomel electrode, this is calculated by considering the potential difference between the calomel electrode in standard condition and in a saturated KCl solution.
Explanation:Firstly, the difference in electrode potential between calomel electrode in standard conditions and in a saturated KCl solution is (E ° = 0.268 V) - (E (saturated KCl) = 0.241 V) = 0.027 V. This difference indicates the change in potential when the electrode is in a saturated KCl solution compared to the standard conditions. Given that the E ° for the silver-silver chloride electrode is 0.222 V in standard conditions, the potential of the Ag/AgCl electrode in a saturated KCl solution can be predicted as E (Ag/AgCl, saturated KCl) = (E ° Ag/AgCl) - (ΔE Calomel) = 0.222 V - 0.027 V = 0.195 V. This implies that the potential of the silver-silver chloride electrode in a saturated KCl solution should be 0.195 V.
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Which of the following numerical expressions gives the number of particles in 2.0 g of Ne?
A. 6.0 x 10^23 particles/mol / 2.0 g
B. 6.0 x 10^23 particles/mol / 20.18 g/mol
C. 2.0 g (6.0 x 10^23 particles/mol) / 20.18 g/mol
D. 20.18 g/mol (6.0 x 10^23 particles/mol) / 2.0 g
Answer:
Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol
Option C is correct
Explanation:
Step 1: Data give
Mass of Ne = 2.0 grams
Molar mass of neon = 20.18 g/mol
Number of Avogadro = 6.0 *10^23 /mol
Step 2: Calculate number of moles of neon
Moles Ne = Mass of ne / Molar mass of ne
Moles Ne = 2.0 / 20.18 g/mol
Moles Ne = 0.099 moles
Step 3: Calculate nulber of particles
Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22
Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)
Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol
Option C is correct
Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above?A) Decreasing the temperature B) Increasing the temperature C) Decreasing the volume of the reaction vessel D) Increasing the volume of the reaction vessel
Answer:
Explanation:
You did not provide the reaction. However, you should know that only change in temperature affects the value of an equilibrium constant (Keq) by the equation, K=Ae>-RT
Answer:
B) Increase in temperature
Explanation:
The equilibrium constant and temperature are inversely proportionate. So, if one increases the other has to decrease.
A hypothetical metal is known to have an electrical resistivity of 3.3×10−8(Ω⋅m). A current of 25 A is passed through a specimen of this metal 15 mm thick. When a magnetic field of 0.95 tesla is simultaneously imposed in a direction perpendicular to that of the current, a Hall voltage of −2.4×10−7V is measured. Compute the following:
(a) the electron mobility for this metal
(b) the number of free electrons per cubic meter
Answer:
a) The electron mobility for this metal is 0.00455 m²V/s
b) There are 4.12 * 10^28 free electrons per cubic meter
Explanation:
Step 1: Data given
electrical resistivity of 3.3 * 10^−8(Ω⋅m)
A current of 25 A is passed through a specimen of this metal 15 mm thick.
A magnetic field of 0.95 tesla.
A Hall voltage of −2.4×10−7V is measured
Step 2: Calculate the conductivity
σ= 1/ ρ
⇒ with ρ = electrical resistivity of 3.3*10^−8Ω*m
σ = 1/3.3*10^−8
σ = 30303030.3 = 3*10^7/ Ω*m
step 3: Calculate the electron mobility
µe = Vdσ / IxBz
⇒ with µe = the electron mobility in m²V/s
⇒ V = voltage of 2.4*10^−7V
⇒ d = 0.015 meter
⇒ σ = the conductivity = 3*10^7/ Ω*m
⇒ Ix = the current = 25 A
⇒ Bz = magnetic field = 0.95 tesla
µe = (2.4 * 10^-7 * 0.015 * 3*10^7) / (25*0.95)
µe = 0.00455 m²V¨/s
The electron mobility is 0.00455 m²V¨/s
Step 4: Calculate the number of free electrons per cubic meter
n = σ/e*µe
⇒ e= the charge of an electron = 1.6 * 10^-19 Coulombs
⇒ σ = The conductivity = 3*10^7/ Ω*m
⇒ µe = The electron mobility = 0.00455 m²V¨/s
n = 3*10^7/ Ω*m /(1.6 * 10^-19 * 0.00455)
n = 4.12 * 10^28 / m³
There are 4.12 * 10^28 free electrons per cubic meter
The electron mobility of the metal is 3.18 × 10^-3 m^2/Vs. The number of free electrons per cubic meter is 1.52 × 10^29 electrons/m^3.
Explanation:To calculate the electron mobility and the number of free electrons per cubic meter for the given metal, we can use the equations involving Hall effect and resistivity. The formula for electron mobility is given by μ = -VH / (B * I * t), where VH is the Hall voltage, B is the magnetic field, I is the current, and t is the thickness. Substituting the values, we get the electron mobility as 3.18 × 10^-3 m^2/Vs.
The formula for the number of free electrons per cubic meter is given by n = 1 / (e * R), where e is the charge of an electron (1.6 × 10^-19 C) and R is the resistivity. Substituting the values, we get the number of free electrons per cubic meter as 1.52 × 10^29 electrons/m^3.
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You can measure how much of something you have by counting individual objects. For example, you can count the number of cookies in a bag or the number of pages in your notebook. There is a name for a number of atoms, ions, or molecules. One mole of a substance is equal to 6.02 x 10^23 atoms, ions, or molecules of that substance. You can determine the number of moles in a substance by obtaining the mass of the substance.
Objectives
• To measure the masses of common compounds and objects
• To calculate the moles and atoms from the experimental masses
a. Salt(NaCl)
b. Sugar(C6H22O11)
c. Chalk(CaCO3)
d. Sand(SiO2)
e. Water(H2O)
f. Balance
g. Spoon
Answer:
Explanation: first use a balance to get the masses of these items. Once the masses of these items are measured, we can then calculate the moles.
The mole is = mass/molarmass
The common laboratory solvent diethyl ether (ether) is often used to purify substances dissolved in it. The vapor pressure of diethyl ether, CH3CH2OCH2CH3, is 463.57 mm Hg at 25°C.
1. In a laboratory experiment, students synthesized a new compound and found that when 21.47 grams of the compound were dissolved in 233.8 grams of diethyl ether, the vapor pressure of the solution was 455.55 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?
Answer: 386.0 g/mol
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : 21.47 g of compound X is present in 233.8 g of diethyl ether
moles of solute (X) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{21.47g}{Mg/mol}[/tex]
moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{233.8g}{74g/mol}=3.160moles[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]\frac{463.57-455.55}{463.57}=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]0.017301=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]M=386.0g/mol[/tex]
The molecular weight of this compound is 386.0 g/mol