The correct answer is slow
Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
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2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
Answer:
Total 5 significant digits.
Explanation:
Significant digits are the numbers that give a meaningful contribution. For example, digit 013 has the 2 significant digits and zero is not a significant digit because digits 1 and 3 give meaningful contribution but digit zero does not value meaningful contribution. Similarly, the 89015 has a total of 5 significant digits and these digits are the 8, 9, 0, 1, and 5.
