Question

1.     2.35 x 10²³ molecules of CO₂ gas are in a sealed container. What is the volume, in liters, of the container?

2.     How do the following affect the position of equilibrium in the following reaction?

     CaC₂(s) + 2H₂O(g) ⇌ Ca(OH)₂(aq) + C₂H₂(g)

             a. addition of acetylene (C₂H₂) gas

             b. removal of some of the water

             c. increase in pressure

            d. removing some of the calcium hydroxide

             e. increase in volume of reaction container

             f. addition of more calcium carbide (CaC₂)

For problem 1, you are required to show ONLY the last 2 steps - the answer and the step leading up to it. For example:

a). 23 L + 51 L = 74 L   b). 15.00 g /25.00 mL = 0.60 g/mL

Answer 1

23 23 -0.39 mole 3 x 22.4L molt - 8.74 L 2 . I male have 6.0g9x 10 molecules. Number of males in 2.35 x 10 molecules At STP Imole gas occupies 22-4 L Volume of 0.390 male occupites gigry 0.39 mole Addition of (C2HQ) shift equilibrium towards Increase in volume number of gaseous molecules Per unit volume decreases, reaction shift on that side which have more number of gastoly J. 2.35x10 23 6.099 x 1023 Calo (s) + 9 H2O(g) Ž Ca (OH)2 (aq.) + C₂H2(g) left. b. Removal of Hoo shift equilibrium towards left Increase pressure equilibrium shift on that side where cless number of gaseous molecules that is towards reight side. di Removing Ca (OH)2 would not effect position of equilibrium as it is not present in gaseous phase e. C. molecules that is left side. Addition of Calg(s) would not effect position of equilibrium. at