The balanced chemical equation is :
5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)
Here the oxidation number of P changed from 0 to -3 in PH₃ and increases from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.
Thus clearly both reduction and oxidation are taking place.
Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.
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<h3>
Answer:</h3>
1.93 g
<h3>
Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
Answer:
Option B. 3.0 M
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity can simply be defined as the mole of solute per unit litre of the solution. Mathematically, it can be expressed as:
Molarity = mole of solute /Volume of solution
With the above formula, we can obtain the molarity of the solution as follow:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity = mole of solute /Volume of solution
Molarity = 9 / 3
Molarity = 3 mol/L = 3.0 M
Thus, the molarity of the solution is 3 M
<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
The oxidation state of Hydrogen in reactant side is 0 and that in product side is +1. Hence the reaction is oxidation.
