Answer:
<h2>A. The image shows Boyle's Law because volume decreases as pressure increases.</h2>
Explanation:
According to Boyle's Law, at constant temperature, pressure and volume have inverse relation that is as pressure would increase, volume would decrease.
According to Charles Law, at constant pressure, the volume is directly proportional to temperature.
According to the given graph and diagram, the temperature is constant and volume and pressure have inverse relation.
Thus, option A is correct.
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Answer:
The potential wrt. calomel is 1.254 V
Explanation:
Given:
Potential wrt. silver chloride 
V
Potential wrt. saturated silver chloride 
V
Potential wrt. SCE 
V
Now potential wrt. hydrogen is given by,
V
And we find for potential wrt. calomel,
potential wrt. hydrogen + potential wrt. SEC
V
Therefore, the potential wrt. calomel is 1.254 V
Answer:
a) yes, it was an hydrate
b) the number of waters of hydration, x = 6
Explanation:
a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.
b) NiCl2. xH2O
mass if dehydrated NiCl2 = 2.3921 grams
mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.
NiCl2.xH2O
mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole
mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole
Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each
for NiCl2 = 0.01846/0.01846 = 1
for H2O = 0.11072/0.01846 = 5.9976 = 6
thus the hydrated sample was NiCl2. 6H2O
