when the top of the pencil moves in one direction, the bottom end on the ground will move in another, but due to the frictional force it will move by a lesser amount, and the center of mass of the pencil will be accelerated in the horizontal direction due to the force of friction

4 0

2 years ago

0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reache

astra-53 [7]

Answer:

10.32g

Explanation:

First thing's first, lets bring out the balanced chemical equation representing the reaction.

CaCO3(s) ⇄ CaO(s) + CO2(g)

From the question;

reaction constant K = [CO2(g)] = 0.220 atm.

To obtain number of moles of CO2, using ideal gas fomula, we have;

n = PV/RT = 0.220*10 / (0.08206*385) = 0.0696(mol)

From the stoichiometry of the equation above;

This is the amount of CaCO3 which has been converted to CaO before additional 0.230 atm CO2(g) is added.

After the additional 0.230 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm. Therefore all this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.230*10.0/(0.08206*385) = 0.0728 (mol)

Hence the total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0696 + 0.0728 (mol) = 0.1032 mol

Mass = Number of moles * Molar mass

Mass = 0.1032 * 100 = 10.32g

7 0

3 years ago