Answer:
B. as a food preservative in the manufacture of detergents
.........
Answer:
the answer is D. Because the 1 atm pressure of water is 40.65 or 40.7.
When the block of iron is placed in water the volume of water that is displaced is 27.0 cm³
<u><em> calculation</em></u>
The volume water that is displaced is equal to volume of block of the iron
volume of block of iron = length x width x height
length= 3 cm
width = 3 cm
height = 3 cm
volume is therefore = 3 cm x 3 cm x 3 cm = 27 cm³ therefore the volume displaced = 27 cm³ since the volume of water displaced is equal to volume of block.
The answer to this question I think would be 2: Better
Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.