A gas has an initial volume of 15L and the pressure is at 2atm. If the temperature increases from 330K to 462K and the pressure reduced to 1 ATM, what is the new volume of the gas?

42L

Explanation:

Using the combined gas law equation as follows:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in this question, P1 = 2atm, V1 = 15L, T1 = 330K, T2 = 462K, P2 = 1atm, V2 = ?

Using the formula;

P1V1/T1 = P2V2/T2

2 × 15/330 = 1 × V2/462

30/330 = V2/462

Cross multiply;

462 × 30 = 330 × V2

13860 = 330V2

V2 = 13860 ÷ 330

V2 = 42L

The new volume is 42L

it takes on an electronical energy

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