Answer:
Carbon dioxide and hydrogen monoxide
Answer:
150.1 mL
Explanation:
Step 1: Given data
- Density of benzene (ρ): 0.879 g/mL
- Mass of the sample of benzene (m): 131.9 g
- Volume of the sample of benzene (V): ?
Step 2: Calculate the volume of the sample of benzene
Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.
ρ = m/V
V = m/ρ
V = 131.9 g/(0.879 g/mL)
V = 150.1 mL
Answer:
Increase pressure 3X => increase Temperature 3X
Explanation:
Gay-Lussac Law => T ∝ P => T =kP => Empirical Relationship => T₁/P₁ = T₂/P₂
=> T₂ = T₁P₂/P₁
Given P₂ = 3P₁ => T₂ = T₁(3P₁)/P₁ = 3T₁
Answer:
392.97 litres
Explanation:
From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.
At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals
356.16L.
Now, we can use the general gas equation to get the volume produced at the values given.
We have the following values;
V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm
The general form of the general gas equation is given as :
(P1V1)T1 = (P2V2)/T2
After substituting the values , we get V2 to be 392.97Litres
