Answer:
the final volume of the gas is
= 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure
:


The workdone W =
V
The change in volume ΔV= 
ΔV = 
ΔV = 
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is 

multiply through by (-), we have:

= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is
= 1311.5 mL
Answer:
Mass is the measure of how much matter is in an object
Explanation:
˙˚ʚ(´◡`)ɞ˚˙
Mass of Oxygen (O₂) : = 88.16 g
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
C₆H₁₄+ O₂ → CO₂ + H₂0
25 g C₆H₁₄
Required
mass of oxygen (O₂)
Solution
Balanced equation
2C₆H₁₄ + 19O₂ ⇒12 CO₂ + 14 H₂O
mol C₆H₁₄ (MW=86,18 g/mol) :
= mass : MW
= 25 g : 86.18 g/mol
= 0.29
From the equation, mol ratio of C₆H₁₄ : O₂ = 2 : 19, so mol O₂ :
= 19/2 x mol C₆H₁₄
= 19/2 x 0.29
= 2.755
Mass O₂(MW=32 g/mol) :
= 2.755 x 32
= 88.16 g