Answer:
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate.
Explanation:
Van Deemter equation in chromatography, relates the variance per unit length of a separation column to the linear mobile phase velocity by considering physical, kinetic, and thermodynamic properties of a separation.
The formula is:
H = A + B/u + C×u
Where:
H = HETP (plate height)
A = Eddy diffusion term .
B = Longitudinal diffusion.
C = Resistance against mass transfer.
u = Linear velocity .
A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow. TRUE. Eddy difussion term depends of column packing
B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. TRUE. Longitudinal difussion is another term that is specific to a column. Also, as the formula is H = A + B/u + C×u, the impact of B on the plate height is inversely proportional to the linear velocity.
C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity. TRUE. C is the time that system needs to equilibirum. Based on the formula, the impact of C on plate height is proportional to the linear velocity
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate. FALSE. The optimal flow is the minimum of the graph
E) Linear velocity is the flow rate of the mobile phase. TRUE. The Van Deemter equation describes u as the linear velocity of the mobile phase.
I hope it helps!
Final answer:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
Explanation:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
This means that in order to minimize the plate height, the optimal flow rate is not at the maximum point on the plot of plate height versus flow rate. The optimal flow rate is usually at a point of minimum plate height.
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°.
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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The ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is first order in C5H10 and first order in O3.
1.Complete the rate law for this reaction in the box below.
Use the form k[A]m[B]n... , where '1' is understood for m, n ...(don't enter 1) and concentrations taken to the zero power do not appear.
Rate =
2.In an experiment to determine the rate law, the rate of the reaction was determined to be 649 Ms-1 when [C5H10] = 0.128 M and [O3] = 4.41E-2 M. From this experiment, the rate constant is ______M-1s-1.
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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The list that correctly indicates the order of metallic character is __________.
A) B > N > C
B) F > Cl > S
C) Si > P > S
D) P > S > Se
E) Na > K > Rb
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.
Which of the slightly soluble salts below will be more soluble in acidic solution than in pure water?
a. Mg(OH)2
b. CaCO3
c. AgCl Mg(OH)2
d. only Mg(OH)2,
e. CaCO3, and
f. AgCl
g. CaCO3 only
h. both Mg(OH)2 and CaCO3
Answer:
h. both Mg(OH)₂ and CaCO₃
Explanation:
Let's consider the solution of Mg(OH)₂ according to the following equation:
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
In acidic solution, OH⁻ reacts with H⁺ to form H₂O.
OH⁻(aq) + H⁺(aq) ⇄ H₂O(l)
According to Le Chatelier's principle, since [OH⁻] decreases, the solution of Mg(OH)₂(s) shifts toward the right, increasing its solubility.
Let's consider the solution of CaCO₃ according to the following equation:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
In acidic solution, CO₃²⁻ reacts with H⁺ to form HCO₃⁻.
CO₃²⁻(aq) + H⁺(aq) ⇄ HCO₃⁻(aq)
According to Le Chatelier's principle, since [CO₃²⁻] decreases, the solution of CaCO₃(s) shifts toward the right, increasing its solubility.
Let's consider the solution of AgCl according to the following equation:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Cl⁻ does not react with H⁺ because it comes from a strong acid (HCl). Therefore, the solubility of AgCl(s) is not affected by the pH.
Mg(OH)₂ and CaCO₃ are more soluble in acidic solutions due to reactions with H+ ions that remove the OH- and CO₃²⁻ from the solution, driving the dissolution equilibrium forward.
Explanation:The question asks which of the slightly soluble salts listed will be more soluble in acidic solution than in pure water. Specifically, salts like Mg(OH)₂ (magnesium hydroxide) and CaCO₃ (calcium carbonate) will be more soluble in an acidic solution. This is because the acid in the solution will react with the anionic part of the salt, which in the case of Mg(OH)2 is OH- and for CaCO₃ is CO₃²⁻. For example, in an acidic solution, H+ ions will react with OH- to form water, which effectively removes OH- from the solution and drives the dissolution equilibrium forward, increasing the solubility of Mg(OH)₂. Similarly, H+ ions will react with CO₃²⁻ to form HCO₃- (bicarbonate) or even further to H₂CO₃ (carbonic acid), which are more soluble than the carbonate ion, hence increasing the solubility of CaCO₃.
As for AgCl (silver chloride), it will also be more soluble in acidic solution because the chloride ion is not basic, and it does not react with H+ to form a weaker acid. Therefore, the correct answers that are more soluble in acidic solution than in pure water are Mg(OH)₂ and CaCO₃.
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.
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.
For many purposes we can treat propane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a sample of propane gas is raised from to , and at the same time the pressure is changed. If the initial pressure was and the volume increased by , what is the final pressure? Round your answer to the correct number of significant digits.
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
Percent Ionization
Percent ionization for a weak acid (HA) is determined by the following formula:
Percent ionization=[HA] ionized[HA] initial×100%
For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization.
A certain weak acid, HA, has a Ka value of 7.6×10−7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution.
Express your answer as a percent using two significant figures.
%
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Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.
%
Answer:
(a) 0.26 % (b) 0.80 %
Explanation:
(a)
Given that:
[tex]K_{a}=7.6\times 10^{-7}[/tex]
Concentration = 0.10 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.10&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.10-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]
[tex]7.2\times 10^{-7}=\frac{x^2}{0.10-x}[/tex]
[tex]7.2\left(0.10-x\right)=10000000x^2[/tex]
Solving for x, we get:
x = 0.00026 M
Percentage ionization = [tex]\frac{0.00026}{0.10}\times 100=0.26 \%[/tex]
(b)
Concentration = 0.010 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.010&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.010-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]
[tex]7.2\times 10^{-7}=\frac{x^2}{0.010-x}[/tex]
[tex]7.2\left(0.010-x\right)=10000000x^2[/tex]
Solving for x, we get:
x = 0.00008 M
Percentage ionization = [tex]\frac{0.00008}{0.010}\times 100=0.80 \%[/tex]
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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2. Perception:
a. is faulty.
b. is not always accurate.
c. does not always reflect reality.
d. All of these choices.
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.
At shipper’s dock you are given a shipping paper for 100 cartons of battery acid. You already have 100 pounds of dry Silver Cyanide on board. What precautions do you have to take?
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.
The half life for the decay of carbon-14 is 5.73 x 10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 53.Bq. The activity in a similar-sized sample of fresh wood is measured to be 59.Bq.
1. Calculate the age of the artifact. Round your answer to 2 significant digits.
Final answer:
To find the age of the artifact using carbon-14 activity, we calculate the ratio of current activity to original activity, use the decay formula with the half-life of carbon-14, and solve for time, yielding an estimated age of approximately 888 years.
Explanation:
To calculate the age of the artifact based on its carbon-14 activity, we use the concept of radioactive decay and the half-life of carbon-14. The half-life is the time taken for half of the radioactive atoms in a sample to decay. In the case of carbon-14, its half-life is 5,730 years.
We start by finding the ratio of the current activity to the original activity.
Original activity (fresh wood) = 59 Bq
Current activity (artifact) = 53 Bq
Ratio (current/original) = 53 Bq / 59 Bq = 0.898
Next, we use the decay formula: N(t) = N0 × (1/2)(t/T), where:
N(t) is the remaining amount of substance after time t
N0 is the original amount of substance
T is the half-life of the substance
t is the time that has passed
Plugging in the values we know: 0.898 = (1/2)(t/5730)
To solve for t, we take the natural logarithm of both sides of the equation:
ln(0.898) = ln((1/2)(t/5730))
ln(0.898) = (t/5730) × ln(1/2)
After calculating:
t = 5730 × ln(0.898) / ln(1/2) = 5730 × -0.107 / -0.693
t ≈ 888 years (rounded to two significant digits)
Therefore, the age of the artifact is approximately 888 years old.
Which of the following units for distance is an Sl unit?
1)Centimeter
2)Foot
3)Inch
4) yard
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
Express the following numbers in scientific notation:
a. 60200 L
b. 4520 J
c. 0.00600 mg
d. 0.01023 km
e. 80.70 mL
Answer :
(a) The number in scientific notation is, [tex]6.0200\times 10^4L[/tex]
(b) The number in scientific notation is, [tex]4.520\times 10^3J[/tex]
(c) The number in scientific notation is, [tex]6.00\times 10^{-3}mg[/tex]
(d) The number in scientific notation is, [tex]1.023\times 10^{-2}km[/tex]
(e) The number in scientific notation is, [tex]8.070\times 10^1mL[/tex]
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as [tex]5.0\times 10^3[/tex]
889.9 is written as [tex]8.899\times 10^{-2}[/tex]
In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex] and [tex]8.899\times 10^{-2}[/tex] are written in the scientific notation.
[tex]8.89\times 10^{-2}[/tex] this is written in the scientific notation and the standard notation of this number will be, 0.00889.
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
(a) As we are given the 60200 L in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 60200L=6.0200\times 10^4L[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]6.0200\times 10^4L[/tex]
(b) As we are given the 4520 J in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 4520J=4.520\times 10^3J[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]4.520\times 10^3J[/tex]
(c) As we are given the 0.00600 mg in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 0.00600mg=6.00\times 10^{-3}mg[/tex]
As, the decimal point is shifting to right side, thus the power of 10 is negative.
Hence, the correct answer is, [tex]6.00\times 10^{-3}mg[/tex]
(d) As we are given the 0.01023 km in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 0.01023km=1.023\times 10^{-2}km[/tex]
As, the decimal point is shifting to right side, thus the power of 10 is negative.
Hence, the correct answer is, [tex]1.023\times 10^{-2}km[/tex]
(e) As we are given the 80.70 mL in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 80.70mL=8.070\times 10^1mL[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]8.070\times 10^1mL[/tex]
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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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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How many electrons are transferred in the following reaction?
(The reaction is unbalanced.)
Mg(s) + Al3+(aq) → Al(s) + Mg2+(aq)
Answer:
6
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]Mg(s)+Al^{3+}(aq.)\rightarrow Al(s)+Mg^{2+}(s)[/tex]
The half cell reactions for the above reaction follows:
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-[/tex]
Reduction half reaction: [tex]Al^{3+}+3e^-\rightarrow Al[/tex]
Magnesium is loosing 2 electrons to form the magnesium cation. Thus, it is getting oxidized. Aluminum anion is gaining 3 electrons to form Aluminum. Thus, it is getting reduced.
Thus, balancing the half-reactions as:-
Oxidation half reaction: [tex]3Mg\rightarrow 3Mg^{2+}+6e^-[/tex]
Reduction half reaction: [tex]2Al^{3+}+6e^-\rightarrow 2Al[/tex]
Thus, total number of electrons transferred = 6
Answer:
3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al
In this reaction 6 electrons are transferred
Explanation:
Step 1: The half reactions
Mg -2e- ⇆ Mg^2+
Al^3+ +3e- ⇆ Al
Step 2: Balance both equations
3*(Mg -2e- ⇆ Mg^2+)
2(Al^3+ +3e- ⇆ Al)
3Mg -6e- ⇆ 3Mg^2+
2Al^3+ +6e- ⇆ 2Al
Step 3: The netto reaction
3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al
In this reaction 6 electrons are transferred
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
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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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 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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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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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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A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?
Answer:
a) 231.9 °C
b) 100% Sn
c) 327.5 °C
d) 100% Pb
Explanation:
This is a mixture of two solids with different fusion point:
[tex]Tf_{Pb}=327.5 C[/tex]
[tex]Tf_{Sn}=231.9 C[/tex]
Given that Sn has a lower fusion temperature it will start to melt first at that temperature.
So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.
The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.
At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.
Which of the following is a definition of glass transition temperature?
The temperature at which a glass transforms from a solid to liquid.
The temperature at which a glass transforms from a liquid to solid.
The temperature at which there is a slight decrease in the slope of the temperature versus specific volume curve.
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.
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.
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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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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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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