PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) For the reaction at equilibrium, if the volume of the container is increased, the amount of PCl5 present will

a.
decrease.

b.
stay the same.

c.
triple.

d.
increase.

e.
double.

If 0.20 mole of Cl 2 is added to the equilibrium mixture, will the concentration of PCl 5 increase or decrease? If the equilibrium constant for this reaction is expressed in terms of concentrations, written as follows: Here, , and represent the molar concentrations of, , and respectively. Chapter 10, Problem 60CQ is solved.

Step-by-step explanation:

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3