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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Answer:
Above all are correct answer.
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1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = <span>1043.77 grams Li(NO3)2</span>
The two samples don’t contain different atoms so it will be false
Answer:
10.10
Explanation:
Step 1: Write the basic dissociation reaction for pyridine
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.9 × 10⁻⁹
Step 2: Calculate [OH⁻]
For a weak base, we will use the following expression.
[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M
Step 3: Calculate pOH
We will use the definition of pOH.
pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9
Step 4: Calculate pH
We will use the following expression.
pH = 14 - pOH = 14 - 3.9 = 10.10