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hram777 [196]
3 years ago
6

You could add aquous hydrochloric to the solution to precipitate out silver chloride. What volume of a 0.130 M aqueous hydrochlo

ric solution is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution? You could add solid potassium chloride to the solution to precipitate silver chloride. What mass of potassium chloride is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution?
Chemistry
1 answer:
Mila [183]3 years ago
5 0
1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.

2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
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95.6 cal
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Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
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m
=
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c
Pb
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0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
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∘
C
Δ
T
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∘
C
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q
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q
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(
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⋅
∘
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×
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24.6
∘
C
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(rounded to three significant figures)
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