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.
To know more about oxidation number here:
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Answer:
Explanation:
Option A is the correct answer
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
Answer:
V ∝ abc
Explanation:
This task is a joint variation task involving only direct proportionality:
Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.
Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.
In this exercise:
If a ∝ b and a ∝ c, then a ∝ bc
Taking the above three proportionalities,
V ∝ a ∝ b ∝ c
V ∝ a ∝ bc
V ∝ abc
