Answer:
0.0933 moles/Litre
Explanation:
We assume that the number of moles of N- used is equal to the number of moles of Nitrogen containing compounds that are generated due to the fact that the nitrogen containing compound that are produced contain only one nitrogen in each atom. As such, finding the amount of nitrogen used up explains the amount of compound formed. This can be expressed as follows:
Energy cost = 
Given that:
Energy = 100 W for 60 minutes
100 W = 100 J/s
= 100 J/s × (60 × 60) seconds
= 3.6 × 10⁵ J
Let now convert 3.6 × 10⁵ J to eV; we have:
= ( 3.6 × 10⁵ × 6.242 × 10¹⁸ )eV
= 2.247 × 10²⁴ eV
So, number of N-atom used up to form compounds will now be:
= 2.247 × 10²⁴ eV × 
= 1.123 × 10²³ N-atom
To moles; we have:
= 
= 0.186 moles
However, we are expected to leave our answer in concentration (i.e in moles/L)
since we are given 2L
So; 0.186 moles ⇒ 
= 0.0933 moles/Litre
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:
A white precipitate of magnesium hydroxide is formed in the above reaction.
Ionic form of the above equation follows:
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
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