Answer:
79.04 L
Explanation:
We are given;
Initial Volume; V1 = 6.24L
Initial Pressure; P1 = 760 mm Hg
Final pressure; P2 = 60.0mm Hg
To solve for final volume, we will use Boyles law;
P1•V1 = P2•V2
Let's make V2 which is the final volume the subject;
V2 = (P1•V1)/P2
V2 = (760 × 6.24)/60
V2 = 79.04 L
Answer:
The equivalent weight of M is approximately 31.8 g
The equivalent weight of N is approximately 27.98 g
Explanation:
The given parameters are;
The percentage of the the metal M in in the chloride = 47.25%
Where by the chemical formula for the metal chloride is MClₓ, we have;
47.25% of the mass of MClₓ = Mass of M = W
Therefore, we have;

0.4725 × (W + 35.5·x) = W
0.4725·W + 0.4725×35.5×x = W
W - 0.4725·W = 16.77·x
0.5275·W = 16.77·x
W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M
The equivalent weight of M = 31.799 ≈ 31.8 g
Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g
The equivalent weight of N = 27.98 g.
Answer:
The relevant equation is:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Explanation:
1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.
The formed CO₂ is the reason why you noticed bubbles as the reaction took place
Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
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The answer to the question is letter b