Each water molecule consists of two atoms of the element hydrogen joined to one atom of the element oxygen. An interesting property of water is the ability of its molecules to “stick together.” This occurs because one side of each water molecule is slightly negative and the other side is slightly positive. The positive portion of a water molecule is attracted to the negative portion of an adjacent water molecule. As a result, water molecules are called polar molecules. They attract other water molecules like little magnets. It is most likely ionic bonding but between hydrogen and oxygen it is covalent.
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Endothermic reactions. These are reactionsthat take in energy from the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surroundings to get colder
Answer:
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Explanation: From the periodic tables we can drive elements with the electronic configuration
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
