The density of the liquid is 1.027g/cm³
D=M/V
D=46.83/45.6
D=1.027g/cm³
Answer:
it is 50
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The 3 and 2 to the right of the components are subscriptions.
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