24.568 cm of chlorine at 25° C will occupy what volume
at -25°C while the pressure remains constant?

20.446cm3

Explanation:

The following were obtained from the question given:

V1 (initial volume) = 24.568 cm3

T1 (initial temperature) = 25°C = 25 + 273 = 298K

T2 (final temperature) = - 25°C = - 25 + 273 = 248K

V2 (final volume) = ?

Since the pressure is constant, the gas is obeying Charles' law. With the Charles' law equation V1/T1 = V2/T2, we can easily calculate the final volume as follow:

V1/T1 = V2/T2

24.568/298 = V2/248

Cross multiply to express in linear form as shown below:

298 x V2 = 24.568 x 248

Divide both side by 298

V2 = (24.568 x 248) /298

V2 = 20.446cm3

From the calculations made above, the volume of the gas at -25°C is 20.446cm3

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2