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Mariana [72]
4 years ago
14

A decrease in the amount of a gas in a container may mean a(n) __________. increase in pressure increase in volume decrease in p

ressure decrease in volume
Chemistry
1 answer:
enot [183]4 years ago
5 0
A decrease in the amount of gas in a container will mean a decrease in pressure.
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Two processes are described below:
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Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subj
denis23 [38]

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

3 0
4 years ago
Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years
Sliva [168]

Answer : The time passed in years is 2.83\times 10^3\text{ years}

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = 71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}

t=2831.00\text{ years}=2.83\times 10^3\text{ years}

Therefore, the time passed in years is 2.83\times 10^3\text{ years}

8 0
3 years ago
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