Answer:
5 g
Explanation:
The heat required to vaporize ice is the sum of
i) Heat required to melt ice at 0°C
ii) Heat required to raise the temperature from 0°C to 100°C
iii) Heat required to vaporize water at 100°C
Thus;
H = nLfus + ncθ + nLvap
H= n(Lfus + cθ + Lvap)
Lfus = 6.01 kJ/mol
Lvap = 41 kJ/mol
c = 75.38
n =?
2100 = n(6.01 + 75.38(100) + 41)
n = 2100 KJ/7585.01 kJ/mol
n = 0.277 moles
Mass of water = number of moles * molar mass
Mass of water = 0.277 moles * 18 g/mol
Mass of water = 5 g
The answer is B the highest point
Answer: 50. 4g
Explanation:
First calculate number of moles of aluminium in 38.8g
Moles = 38.8g/ 26.982mol/g
= 1.44mol
By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.
4 = 2
1.4 = x
Find the value of x
x= (1.4×2)/4= 0.72 mol
0.72 moles of aluminium oxide are produced from 38.8g of aluminium
Now find the mass of aluminium produced.
Mass = moles × molar mass
= 0.72mol × 69.93 mol/g
= 50.4g
Answer:
b) Delta S < 0
Explanation:
The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).
- If Δn(g) > 0, the entropy increases (ΔS > 0).
- If Δn(g) < 0, the entropy decreases (ΔS < 0).
- If Δn(g) = 0, there is little or no change in the entropy
Let's consider the following equation.
2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)
Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.
