Answer:
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Explanation:
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.
To begin calculating, there is one thing you need to remember :
Then we have
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>
I'm sure it will help.
Answer:
B. 214.02
Explanation:
1 mol of water weighs 18.015 gm and contains 6.023 × 10²³ molecules
From question, We have 7.15 × 10²⁴ molecules
Dividing we get (7.15 × 10 ²⁴) ÷ ( 6.023 × 10²³) = 11.871 molecules
Now, Weight of water = 11.871 × 18.015 = 213.85 which is nearer to option B
