A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is doubled, and the temperature increases to 400 K

The answer to your question is Volume = 600 ml

Explanation:

Data

Volume 1 = 810 ml

Temperature 1 = 270°K

Pressure 1 = 1 atm

Volume 2 = ?

Pressure = 2 atm

Temperature 2 = 400°K

Process

1.- To solve this problem, use the Combine gas law.

               P1V1 / T1 = P2V2 / T2

- Solve for V2

                V2 = P1V1T2 / T1P2

2.- Substitution

                 V2 = (1)(810)(400) / (270)(2)

3.- Simplification

                 V2 = 324000 / 540

4.- Result

                 V2 = 600 ml

600mL

Explanation:

The following were obtained from the question:

V1 = 810mL

P1 = 1atm

T1 = 270K

V2 =?

P2 = 2atm(since the pressure is doubled)

T2 = 400K

The new volume can be obtained by doing the following :

P1V1/T1 = P2V2/T2

1 x 810/270 = 2 x V2 /400

Cross multiply to express in linear form

270 x 2 x V2 = 810 x 400

540 x V2 = 324000

Divide both side by 540

V2 = 324000/540

V2 = 600mL

The new volume of the gas is 600mL

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