A bag of soil will weigh in kilograms
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:
Isotopes
Explanation:
Isotopes are atoms of the same element that contain an identical number of protons, but a different number of neutrons. Despite having different numbers of neutrons, isotopes of the same element have very similar physical properties.
Answer:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.