8
It’s 8 bc I said it was 8 ;)
Answer:
3.18 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 0.985 atm
- Initial volume (V₁): 3.65 L
- Final pressure (P₂): 861.0 mmHg
Step 2: Convert P₁ to mmHg
We will use the conversion factor 1 atm = 760 mmHg.
0.985 atm × 760 mmHg/1 atm = 749 mmHg
Step 3: Calculate the final volume of the gas
Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L
Its okay my friend. you dont need to over stress it.
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.
