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likoan [24]
3 years ago
12

Convert 15.2 moles of K to atoms of K.

Chemistry
1 answer:
choli [55]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

Multiply by the given number of moles: 15.2

15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

The moles of potassium cancel.

15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }

The denominator of 1 can be ignored.

15.2 * {6.022*10^{23} \ atoms \ K }{

Multiply.

9.15344*10^{24} \ atoms \ K

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

9.15*10^{24} \ atoms \ K

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

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What factors affect the dynamic state of equilibrium in a chemical reaction and how?
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Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

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For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

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\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

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The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

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For reactions that involve gases,

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However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
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