It is because water is a liquid form.
Answer:
in the presence of water H2O
Na2CO3 (S) --> 2Na+ (aq)+ (CO3)2-(aq)
One mole of sodium carbonate produces two moles of Na+ ions
Therefore 0.207 moles produces 0.414 moles of Na+ ions
= 0.414 moles of Na+ ions
Explanation:
In water
Na2CO3 --> 2Na+ (aq)+ (CO3)2-(aq)
In a limited reaction, the carbonate ion reacts with the water molecules as follows
(CO3)2-(aq) + H2O←→HCO3-(aq) + OH-(aq)
sodium carbonate or soda ash dissolves in water to give 2 sodium cations and one carbonate anion
Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.
Part 1:
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.
Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.
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:
B) microscopic
Explanation:
A scanning tunneling microscope allows imaging of microscopic particles.