Carbon dioxide and water react to form methanol and oxygen, like this: CO2(g) + H2O (g) > CH3OH (l) + O2 (g) At a certain temperature, a chemist finds that 8.6 L a reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition: compound amount CO2 2.25 gH2O 2.72 gCH3OH 3.82 gO2 1.98 gCalculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

68

Explanation:

First, we will calculate the molar concentration (M) of each substance using the following expression.

M = mass of the substance / molar mass of the substance × volume of solution

CO₂

M = 2.25 g / 44.01 g/mol × 8.6 L = 0.0059 M

H₂O

M = 2.72 g / 18.02 g/mol × 8.6 L = 0.018 M

CH₃OH

M = 3.82 g / 32.04 g/mol × 8.6 L = 0.014 M

O₂

M = 1.98 g / 32.00 g/mol × 8.6 L = 0.0072 M

Let's consider the following reaction at equilibrium.

CO₂(g) + H₂O(g) ⇄ CH₃OH(l) + O₂(g)

The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then,

Kc = [O₂] / [CO₂] × [H₂O]

Kc = 0.0072 / 0.0059 × 0.018

Kc = 68

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.