The water will boil at C) 80°C
<h3>Further explanation</h3>
Given
Vapour pressure of water = 47 kPa
Required
Boiling point of water
Solution
We can use the Clausius-Clapeyron equation :

Vapour pressure of water at boiling point 100°C=101.325 kPa
ΔH vap for water at 100°C=40657 J/mol
R = 8.314 J/mol K
T₁=boiling point of water at 101.325 kPa = 100+273=373 K
Input given values :

Answer:
0.80 Moles of Hydrochloric Acid would be required to produce 0.40 Moles of hydrogen gas due to the equation ratios
Explanation:
How many moles of hydrochloric acid are required to produce .40 moles of hydrogen gas?
Hydrochloric acid = HCl
Hydrogen gas = H2
H = 1+
Cl = 1-
HCL = H2 + Cl2
2HCl = H2 + Cl2
2:1:1
0.80 = 0.40 + 0.40
Answer:oxidation number H = +1
oxidation number O = -2
let x = oxidation number Cl
in HClO3
+1 + x -6 =0
x = +5
Explanation:
Answer:
p = 1.5 atm
Explanation:
pV = nRT
p = nRT/V = [1.8 mol×(0.082 atm L/mol K)×247.8 K]/24.5 L
p = 1.5 atm