Answer:
pH = 4.57
Explanation:
pH = pKa + log ([OAc⁺]/[HOAc])
Ka(HOAc) 1.8 x 10⁻⁵ => pKa = -log(1.8 x 10⁻⁵) = 4.74
[OAc⁻] = 0.20M
[HOAc] = 0.30M
pH = 4.74 + log([0.20]/[0.30]) = 4.47 + (-0.17) = 4.57
The percent yield : 81.5%
<h3>Further explanation</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
Reaction
2CO+O₂⇒2CO₂
mol CO(MW=28,01 g/mol)

mass CO₂ (theoretical)(MW=44,01 g/mol)

the percent yield :

Alpha is the -OH group at the anomeric position is down. beta is up
a. alpha
b. beta
c. beta
d. alpha
to draw the other anomer, just flip the OH at the anomeric position
Answer:
101.56 of H₂O
Explanation:
The balanced equation for the reaction is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the mass of CH₄ that reacted and the mass of H₂O produced from the balanced equation. This is illustrated below:
Molar mass of CH₄ = 12 + (4×1.01)
= 12 + 4.04
= 16.04 g/mol
Mass of CH₄ from the balanced equation = 1 × 16.04 = 16.04 g
Molar mass of H₂O = (2×1.01) + 16
= 2.02 + 16
= 18.02 g/mol
Mass of H₂O from the balanced equation = 2 × 18.02 = 36.04g
SUMMARY:
From the balanced equation above,
16.04 g of CH₄ reacted to produce 36.04 g of H₂O.
Finally, we shall determine the mass of water, H₂O produced by the reaction of 45.2 g of methane, CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16.04 g of CH₄ reacted to produce 36.04 g of H₂O.
Therefore 45.2 g of CH₄ will react to produce = (45.2 × 36.04)/16.04 = 101.56 g of H₂O.
Thus, 101.56 of H₂O were obtained.