1 is The first one
<span>2 is the third one
3 is the second one
4 is the 4th one</span>
We are going to use this equation:
ΔT = - i m Kf
when m is the molality of a solution
i = 2
and ΔT is the change in melting point = T2- 0 °C
and Kf is cryoscopic constant = 1.86C/m
now we need to calculate the molality so we have to get the moles of NaCl first:
moles of NaCl = mass / molar mass
= 3.5 g / 58.44
= 0.0599 moles
when the density of water = 1 g / mL and the volume =230 L
∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg
now we can get the molality = moles NaCl / Kg water
=0.0599moles/0.23Kg
= 0.26 m
∴T2-0 = - 2 * 0.26 *1.86
∴T2 = -0.967 °C
Answer: 68.
Explanation:
The atomic number is the number that idenfities an element.
The atomic number is the number of protons of an element.
Every element has a different number of protons.
The elements are arranged in the periodic table as per their atomic number (number of protons).
The first element is hydrogen (H), its atomic number is 1, and it has 1 proton.
The second element is helium (He), its atomic number is 2 (it has 2 protons)-
Those two elements form the first period (row) of the periodic table.
The second row (period) has the elements Li, Be, B, C, N, O, F and Ne, whose respective atomic numbers (number of protons are) 3, 4, 5, 6, 7, 8, 9, and 10, respectively.
And so, you may idenfity each of the 118 elements known, with a different atomic number (number of protons).
Answer:
A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic
B. S(g) + O₂(g) --------> SO₂(g) exothermic
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic
D. 2F(g) ---------> F₂(g) exothermic
Explanation:
The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.
A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy
B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic
D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
