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Basile [38]
3 years ago
14

Which of the following would shift the following equilibrium system to the right? C2H2(g) + H2O(g) ⇌ CH3CHO(g) Removing H2O(g) f

rom the system Adding CH3CHO(g) to the system Removing C2H2(g) from the system Adding H2O(g) to the system
Chemistry
2 answers:
sergey [27]3 years ago
4 0
<h2>Answer:</h2>

<u>The right choice is </u><u>Adding H2O(g) to the system</u>

<h2>Explanation:</h2>

If we dilute a solution by adding solvent like water, all of the concentrations will decrease and a change in Q occurs. If there are more species in solution that are products than reactants then Q will decrease. Hence the reaction will then shift towards the right side which is product side to reach equilibrium because there will be an increase the concentration of the reactants side.

Fittoniya [83]3 years ago
3 0

Answer:

Adding H₂O(g) to the system.

Explanation:

  • Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

<u><em>1) Removing H₂O(g) from the system:</em></u>

  • This will decrease the concentration of the reactants side, so the reaction will be shifted to the left side to suppress the removal of H₂O(g) from the system.

<u><em>2) Adding CH₃CHO(g) to the system :</em></u>

  • This will increase the concentration of the products side, so the reaction will be shifted to the left side to suppress the adding CH₃CHO(g) to the system.

<u><em>3) Removing C₂H₂(g) from the system:</em></u>

  • This will decrease the concentration of the reactants side, so the reaction will be shifted to the left side to suppress the removal of C₂H₂(g) from the system.

<u><em>4) Adding H₂O(g) to the system:</em></u>

  • This will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the addition of H₂O(g) to the system.
  • <u><em>So, it is the right choice.</em></u>
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CaCO3(s) ⟶ CaO(s)+CO2(s) 

<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>

<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed. 
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3 years ago
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Consider the reaction.
N76 [4]

Answer: m = 50 g ZnSO4

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0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

= 0.311 moles ZnSO4

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Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~5
valentinak56 [21]

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

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The molecular mass of ATP is 507.18 g per mol

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mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

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3 years ago
PLEASE HURRY. Which of the molecules below is propyne?
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The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
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Answer:

(a)

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(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

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