Answer:
Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation
Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
the very bottom of the ocean and they are relevant by most of the animals feed off of the animals that come out of he vents
C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.
Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ
By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C
Thus,
-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
<em>Tf = 36.31°C</em>
