Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.
Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric
H
3
O
+
.
Moles of nitric acid:
26.0
×
10
−
3
⋅
L
×
8.00
⋅
m
o
l
⋅
L
−
1
=
0.208
⋅
m
o
l
H
N
O
3
(
a
q
)
.
And, moles of hydrochloric acid:
88.0
×
10
−
3
⋅
L
×
5.00
⋅
m
o
l
⋅
L
−
1
=
0.440
⋅
m
o
l
H
C
l
(
a
q
)
.
This molar quantity is diluted to
1.00
L
. Concentration in moles/Litre =
(
0.208
+
0.440
)
⋅
m
o
l
1
L
=
0.648
⋅
m
o
l
⋅
L
−
1
.
Now we know that water undergoes autoprotolysis:
H
2
O
(
l
)
⇌
H
+
+
O
H
−
. This is another equilibrium reaction, and the ion product
[
H
+
]
[
O
H
−
]
=
K
w
. This constant,
K
w
=
10
−
14
at
298
K
.
So
[
H
+
]
=
0.648
⋅
m
o
l
⋅
L
−
1
;
[
O
H
−
]
=
K
w
[
H
+
]
=
10
−
14
0.648
=
?
?
p
H
=
−
log
10
[
H
+
]
=
−
log
10
(
0.648
)
=
?
?
Alternatively, we know further that
p
H
+
p
O
H
=
14
. Once you have
p
H
,
p
O
H
is easy to find. Take the antilogarithm of this to get
[
O
H
−
]
.
Answer link
Answer:
protons: 36
neutrons:48
electrons:36
Explanation:
the number of protons in an element is = to the atomic # (36)
the number of neutrons is the atomic mass - atomic # (84-36=48)
In a neutral charged element the # of protons = # of electrons
Answer:
The answer to your question is maybe letter D, but the last oxygen needs a number 6.
Explanation:
The empirical formula gives the actual elements that form part of a molecule but not the total numbers.
The molecular formula gives the total number of atoms of each element in a molecule.
We must factor the molecular formula to know if a formula is the empirical formula of that.
A. CH₄ C₂H₆ = 2(CH₃) these are not empirical molecular formulas
B. CH₂O C₄H₆O these are not empirical-molecular formulas
C. O₂ O₃ these are not empirical-molecular formulas
D. C₃H₄O₃ C₆H₈O these are not empirical-molecular formulas
the last oxygen needs a number 6 to be
the answer.
Circulating round the nucleus are the electrons in various orbits of different energy levels. Electrons are negatively charged and represented by the symbol 'e'. In the given image the number of protons are -6. Hence the element in question is Carbon as Carbon has the atomic number 6.
Answer:
silicon tetrachloride or tetrachlorosilane
