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azamat
3 years ago
13

Consider the reaction in chemical equilibrium. cocl2(g) co(g) + cl2(g) which is the correct equation for k?

Chemistry
2 answers:
hichkok12 [17]3 years ago
8 0
K=[CO][Cl^2] / [COCl^2]
Sergeu [11.5K]3 years ago
3 0

Answer:

Keq = \frac{P_{CO}*P_{Cl2}  }{P_{COCl2} }

Explanation:

A reaction is said to be at equilibrium when the rate of consumption of reactants equals the rate of formation of products. This state can be depicted in terms of the equilibrium constant (Keq) which is essentially the ratio of concentration (or pressure for gases) of the products to that of the reactants each of which is raised to an appropriate coefficient defined by the balanced chemical equation.

Keq = \frac{[Products]}{[Reactants]}

The given reaction is:

cocl2(g) ↔ co(g) + cl2(g)

Keq = \frac{P_{CO}*P_{Cl2}  }{P_{COCl2} }

Here the Keq is given in terms of the partial pressure of the gases

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3 years ago
What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?
Aleksandr [31]

Answer:

\boxed {\boxed {\sf 3.3 \ liters}}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

  • molarity= 1.2 mol/L
  • moles of solute= 4.0 mol
  • liters of solution =x

Substitute the values into the formula.

1.2 \ mol/L = \frac{4.0 \ mol}{x}

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}

4.0 \ mol *1=1.2 \ mol/L *x

4.0 \ mol = 1.2 \ mol/L *x

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}

\frac{4.0 \ mol}{1.2 \ mol/L}=x

The units of moles (mol) will cancel.

\frac{4.0 }{1.2 } \ L =x

3.33333333 \ L=x

The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

3.3 \ L\approx x

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