The stopcock connecting a 2.25 l bulb containing oxygen gas at a pressure of 4.00 atm, and a 5.25 l bulb containing neon gas at a pressure of 8.00 atm, is opened and the gases are allowed to mix. assuming that the temperature remains constant, what is the final pressure of the system in atm?

The final pressure of the system is 6.8 atm

Explanation:

Step 1: Data given

The oxygen bulb has a volume of 2.25L and pressure of4.00atm

The neon bulb has a volume of 5.25 L and a pressure if 8.00 atm

Step 2: Calculate the total volume

V1 + V2 = Vtotal

2.25 L + 5.25 L = 7.50 L

So the volume increases to 7.50 liters

Step 3: Calculate the partial pressure of oxygen

P1V1 = P'V'

P' = P1V1/V'

 ⇒with P1 = 4.00 atm

 ⇒ with V1 = 2.25 L

⇒ with P' = TO BE DETERMINED

⇒ with V' = 7.5 L

P' = (4*2.25) / 7.5

P'  = 1.2 atm = partial pressure of oxygen

Step 4: Calculate partial pressure of neon

partial pressure of neon (8.00*5.25)/7.5 = 5.6 atm

Step 5: Calculate final pressure

The total pressure = 1.2atm + 5.6atm = 6.8 atm.

The final pressure of the system is 6.8 atm

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

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