Explanation:
I think B a chemical change
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Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
Answer:
H₂Se
Explanation:
A way of estimating the acidity of a weak acid is by analizing the<em> stability of the formed anion</em>. In this case, we should find a Group 6A element that in its anionic forms (HX⁻ and X⁻²) is more stable than HS⁻ and S⁻², thus it would be more acidic in aqueous solution.
The anionic forms of Se are more stable than the forms of S, similarly to how Br⁻ is more stable than Cl⁻.