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Elis [28]
2 years ago
11

8. The following system contains 20.0 psi each of gases A and B in the end bulbs with an evacuated bulb in the centre. Both stop

cocks are opened at once. What is the new overall pressure in the system?
I literally have no idea how to even start this, I know that the answer is 13.3 psi, but I don't have any clue why or how to get there.

Chemistry
1 answer:
Effectus [21]2 years ago
3 0

Answer:

13.3

Explanation:

Before opening the stopcocks you have 40psi in total.

After that, those 40psi will be divided into 3bulbs, so 40/3=13.3psi for the system.

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A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?
olga nikolaevna [1]

 The volume that  will  be occupied at 735  torr and 57 c  is 23.12 L


 <u><em>calculation</em></u>

  • <u><em> </em></u> At STP   temperature=273 k  and  pressure=760 torr
  • <u><em> </em></u>by use of combined  gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

           T1= 273  K

           V1= 18.5 L

          P2= 735 torr

         T2=  57+273= 330 K

          V2=?

  • by making   V2 the formula of subject

     V2= T2P1V1/P2T1

       V2=  [(18.5L  x 330 k  x 760 torr)/(735 torr x 273 k)]= 23.12  L




5 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

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