A given substance is a white, granular solid at 25 °C that does not conduct electricity. It melts at 750 °C wand the melt conducts electricity. It dissolves easily in water and produces a conductive solution. Identify the type of crystal and justify your answer.

94 ATP molecules

To answer the question we need to understand the beta oxidation of fatty acid.

What happens during oxidation of a fatty acid with even number of carbon, C₂ₙ?

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:

Beta Oxidation of Fatty Acid:

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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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.

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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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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1. The rate law is Rate = k[C5H10][O3].

2. The rate constant is 1.24E6 M-1s-1.

1. The rate law for the ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is first order in C5H10 and first order in O3. Therefore, the rate law can be written as:

Rate = k[C5H10][O3]

2. The rate constant can be calculated using the following equation:

k = Rate / [C5H10][O3]

Substituting the known values into the equation above, we get the following:

k = 649 Ms-1 / 0.128 M * 4.41E-2 M

k = 1.24E6 M-1s-1

Therefore, the rate constant for the ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is 1.24E6 M-1s-1.

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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:

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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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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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.

Answer:

C) Si > P > S .

Explanation:

In the Periodic Table, the metallic character increases from right to left and from top to bottom.

The list that correctly indicates the order of metallic character is  

A) B > N > C . NO. C is in Group 14 and N is in Group 15.

B) F > Cl > S . NO. F and Cl are in Group 17 and S is in Group 16.

C) Si > P > S . YES. Si is in Group 14, P is in Group 15 and S is in Group 16.

D) P > S > Se. NO. Se is below S in the Group 16.

E) Na > K > Rb. NO. Na is above K, which is above Rb in Group 1.

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.

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°.

Answer:

Option a, centimeter

Explanation:

Foot, Inch and Yard are units used in America, UK, and other english-speaking countries.

These countries are currently reviewing the possibility of adapting the measures to the SI. The centimeter is a unit derived from the meter, base unit for measuring length in the SI, and corresponds to the cgs system, where in addition to the centimeter, is constituted by gram and second units

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

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

Perception is an accurate but imperfect process, influenced by expectations, emotions, and selective attention, which can lead to illusions and misinterpretations of the environment.

Understanding Perception

Perception is the process by which we interpret the information our senses provide. While it is a very accurate system, it is not infallible. Our expectations and emotions can influence our perception, sometimes leading to illusions or inaccurate judgments. For instance, when confronted with an optical illusion, our understanding of size, distance, and color might be challenged, resulting in a misinterpreted reality.

Expectations play a significant role in shaping perception. For example, if someone expects to see a certain outcome, they might unconsciously ignore evidence that contradicts their expectation. This is evident in situations where stereotypes or generalizations are held about people or situations, often resulting in biased communication.

Another aspect is selective attention; we tend to perceive and interpret things that are in line with our focus. If our attention is elsewhere, we might not perceive something obvious. Further, perception is a key component during arguments, where people defend their subjective reality instead of considering the objective external environment.

In summary, perception is a complex process influenced by individual biases and attention, often resulting in a personally constructed reality rather than an accurate reflection of the environment.

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.

Answer: (This question is missing some values).

The  combined gas law is used to determine the change in volume, pressure and temperature of gases. It states that the ratio between the pressure-volume product and temperature is a constant.

Mathematically, P1V1/T1 = P2V2/T2, where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, T1 and T2 are the initial and final temperatures in kelvin.

Explanation:

The boiling point of propane is -42°C.

Assuming the initial and final temperatures are 10°C and 25°C respectively; the volume increased by 20%; if the initial pressure = 1atm, final pressure can be found using the equation P1V1/T1 = P2V2/T2.

P1 = 1atm, P2 = ?, V1 = V, V2 = 0.2V, T1 = 10 + 273K = 283K, T2 = 25 + 273K = 298K.

Making P2 subject of formula, P2 = P1V1T2/V2T1

P2 = 1 * V *298/(0.2 V * 283)

P2 = 5.2atm

Answer:

The equation you should use is:

Meaning of the initials of the equation:

INITIAL PRESSURE (P1)

FINAL PRESSURE (P2)

INITIAL VOLUME (V1)

FINAL VOLUME (V2)

FINAL TEMPERATURE (T2)

INITIAL TEMPERATURE (T1)

then the final equation would be:

(P1XV1) / T1 = (P2XV2) / T2

Explanation:

This equation is due to the fact that the gas is considered to be an ideal gas, so when behaving as such the values of "n" which is the number of moles is the same in the initial and final stage as the constant

"R" that has a value of 0.082 (with their respective units) both at the end and at the beginning of the reaction.

By not varying these components of the equation it is unnecessary to put them, since they would cancel themselves.

In the equation we mentioned before, it is necessary that if you want to know the final pressure, that is, P2, you have to clear it, considering that the final equation for this specific exercise is:

P2 = (P1XV1XT2) / V2

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.

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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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.

I hope it helps!

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.

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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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.

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.

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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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.

To know more about  boiling point,

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The glass transition temperature ([tex]T_g[/tex]) is the temperature at which there is a slight decrease in the slope of the temperature versus specific volume curve.

Glass transition temperature ([tex]T_g[/tex]) is defined as the temperature at which there is a slight decrease in the slope of the temperature versus specific volume curve.

Unlike the melting point, which involves a phase change from solid to liquid, the glass transition temperature is a continuous transition where a supercooled liquid transitions into a glassy, rigid state.

This process does not involve a complete change of state but rather the material's internal structure becoming more rigid. At temperatures above [tex]T_g[/tex], the material behaves more like a quasi-equilibrium liquid, whereas below [tex]T_g[/tex], reorganization among structural units ceases, and the material becomes more glass-like in its properties.

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.

Answer:

At the shipper's dock when it already has the 100 cartons of battery acid, 100 pounds of dry silver cyanide should not be loaded together.

Explanation:

The dry silver cyanide and battery acid are list of products that has been restricted from taking together while travelling because of safety reasons. This has been already present in the "Do not load" division 6.1 i.e. materials such as silver cyanide cannot be loaded with acid or any other corrosive material which could react to make hydrocyanic acid. As hydocyanic acid is an oxidant and would be an agent that help in igniting combustibles like wood, paper oil and clothing.

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.

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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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.

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.

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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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.

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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