Moles Cu+2 = M * V
= 0.05 L * 0.011 m
= 0.00055 moles
when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles
and when the Molarity of EDTa = 0.0630 M
∴ Volume of EDTA = moles / Molarity
= 0.00055 / 0.0630
= 0.0087 L = 8.7 L
Answer:
16Ag + 2C2 ---> 4 Ag4C
Explanation:
Ag + C2 ----> AgC2
The valence of Ag is one
Valence of Carbon is four
A balanced equation would be
16Ag + 2C2 ---> 4 Ag4C
(a)
pH = 4.77
; (b)
[
H
3
O
+
]
=
1.00
×
10
-4
l
mol/dm
3
; (c)
[
A
-
]
=
0.16 mol⋅dm
-3
Explanation:
(a) pH of aspirin solution
Let's write the chemical equation as
m
m
m
m
m
m
m
m
l
HA
m
+
m
H
2
O
⇌
H
3
O
+
m
+
m
l
A
-
I/mol⋅dm
-3
:
m
m
0.05
m
m
m
m
m
m
m
m
l
0
m
m
m
m
m
l
l
0
C/mol⋅dm
-3
:
m
m
l
-
x
m
m
m
m
m
m
m
m
+
x
m
l
m
m
m
l
+
x
E/mol⋅dm
-3
:
m
0.05 -
l
x
m
m
m
m
m
m
m
l
x
m
m
x
m
m
m
x
K
a
=
[
H
3
O
+
]
[
A
-
]
[
HA
]
=
x
2
0.05 -
l
x
=
3.27
×
10
-4
Check for negligibility
0.05
3.27
×
10
-4
=
153
<
400
∴
x
is not less than 5 % of the initial concentration of
[
HA
]
.
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
x
2
0.05
−
x
=
3.27
×
10
-4
x
2
=
3.27
×
10
-4
(
0.05
−
x
)
=
1.635
×
10
-5
−
3.27
×
10
-4
x
x
2
+
3.27
×
10
-4
x
−
1.635
×
10
-5
=
0
x
=
1.68
×
10
-5
[
H
3
O
+
]
=
x
l
mol/L
=
1.68
×
10
-5
l
mol/L
pH
=
-log
[
H
3
O
+
]
=
-log
(
1.68
×
10
-5
)
=
4.77
(b)
[
H
3
O
+
]
at pH 4
[
H
3
O
+
]
=
10
-pH
l
mol/L
=
1.00
×
10
-4
l
mol/L
(c) Concentration of
A
-
in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the
[
A
-
]
.
pH
=
p
K
a
+
log
(
[
A
-
]
[
HA
]
)
4.00
=
−
log
(
3.27
×
10
-4
)
+
log
(
[
A
-
]
0.05
)
=
3.49
+
log
(
[
A
-
]
0.05
)
log
(
[
A
-
]
0.05
)
=
4.00 - 3.49
=
0.51
[
A
-
]
0.05
=
10
0.51
=
3.24
[
A
-
]
=
0.05
×
3.24
=
0.16
The concentration of
A
-
in the buffer is 0.16 mol/L.
hope this helps :)
Answer:
B.) 129.9 grams
Explanation:
To find the mass, you need to use the following equation:
Q = mcΔT
In this equation,
-----> Q = energy (J)
-----> m = mass (g)
-----> c = specific heat (J/g°C)
-----> ΔT = change in temperature (°C)
The specific heat of copper is 0.385 J/g°C. Knowing this, you can plug the given values into the equation and simplify to isolate "m".
Q = mcΔT <----- Equation
5000 J = m(0.385 J/g°C)(200 °C - 100 °C) <----- Insert values
5000 J = m(0.385 J/g°C)(100) <----- Subtract
5000 J = m(38.5) <----- Multiply 0.385 and 100
129.9 = m <----- Divide both sides by 38.5
