Answer: C
Explanation:
In endothermic reactions, enthalpy is positive, and in exothermic reactions, enthalpy is negative, So, if enthalpy is positive, then it is an endothermic reaction, and hence is required for the reaction to occur.
For this question the answer is c
<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
<u>Explanation:</u>
We are given:
Heat of vaporization for water = 2257 J/g
Amount of sweat lost = 307 grams
Applying unitary method:
For 1 g of sweat lost, the energy required is 2257 Joules
So, for 307 grams of sweat lost, the energy required will be = 
Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
