Answer: its 15 its none of those
Explanation:
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:

(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
Answer:
0.208mole of CO2
Explanation:
First, let us calculate the number of mole of HC3H3O2 present.
Molarity of HC3H3O2 = 0.833 mol/L
Volume = 25 mL = 25/100 = 0.25L
Mole =?
Mole = Molarity x Volume
Mole = 0.833 x 0.25
Mole of HC3H3O2 = 0.208mole
Now, we can easily find the number of mole of CO2 produce by doing the following:
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
From the equation,
1mole of HC2H3O2 produced 1 mole of CO2.
Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2