Answer:
1- 0.04 M/s.
2- 0.16 M/s.
Explanation:
- For the reaction: 4PH₃ → P₄ + 6H₂.
<em>The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.</em>
where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".
d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".
d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).
<u><em>(a) At what rate is P₄ changing?</em></u>
∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.
∴ <em>rate of P₄ changing = </em>d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.
<u><em>(b) At what rate is PH</em></u>₃<u><em> changing?</em></u>
∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.
∴ <em>rate of PH</em>₃<em> changing = </em>- d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.
Seven valence electrons .
<span>masss % of KBr = mass of KBr/total mass of solution X 100 = 50/100 X 100 = 50 %</span>
Answer: B. Bohr and D. Rutherford
