Answer:
1. B
2.C
3.D
4.Sorry don't this one :(
5.A
Explanation:
please give me at least 20 points ❤
Answer:
1.023 J / g °C
Explanation:
m = 37.9 grams
ΔT = 25.0*C
H = 969 J
c = ?
The equation relating these equation is;
H = mcΔT
making c subject of formulae;
c = H / mΔT
c = 969 J / (37.9 g * 25.0*C)
Upon solving;
c = 1.023 J / g °C
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
