Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
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Use the ICE table approach as solution:
PbSO₄ --> Pb²⁺ + SO₄²⁻
I - 0 0
C - +s +s
E - s s
Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,
mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
mass of nitrogen = 0.602 g N₂
Therefore, 0.602 g of nitrogen will be required for he reaction.