When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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Answer:
4 lead = Pb
2 nitrogen = N
6 oxygen = O
Explanation:
Know the rules of multiplying wth perentheses.
- Hope that helped! Please let me know if you need further explanation.
Answer:Static electricity works because objects which are otherwise "neutral" (in other words, objects with no net charge) can be polarized. An electric field, like one caused by a nearby charged object, can cause the charges inside of a neutral object — the protons and electrons — to move around a tiny bit.
Explanation:
(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 :)
