Question

Quine

請問這題這樣算錯在哪呢@@重看了幾遍還是想不出少了什麼

=0.01875 mol
Mg02416-40
D
(D) 1. If 0.750 grams of magnesium oxide is placed in a coffee-cup calorimeter
and then 100.0mL of 1.00 M HCI is added, the temperature of the HCI solution
increases from 22.8°C to 28.7°C/Based on this information, calculate the
enthalpy change for this reaction per mol of MgO. You may assume that the
specific heat of the solution is 4.20 J/g. K and the density of the HCI solution is
1.00 g/mL.
(A) -2.54 kJ/mol (B) -29.8 kJ/mol (C)-65.3 kJ/mol (D) -134 kJ/mol

0,759 M₂0 =
2
0.73
40
= 0.0 1873 mol
100ml. IM HCI
AT = 5.9°2
求MgO的莫H变化
AH=(106 × 1.00+ 0,73) × 4. 2× (293+5-9)
= 100.23 x 42×298.9= 118616.5J = 118 kJ
118 kJ
0.01873 = 6.293 kJ/20

Answer 1

我就是我

273+5.9那裡要改成5.9才對
實際上是ΔT=(273+28.7)-(273+22.8)=5.9
不用再加273

Quine

請問他提供的比熱那邊單位不是J/g*K嗎 謝謝🙏！