Answer:
New volume is 14.35 mL
Explanation:
When a system of a gas keeps on constant its temperature and number of moles, the pressure is modified indirectly proportional to the volume:
Pressure increased → Volume decreased
Pressure decreased → Volume increased.
The relation you have to apply is: P₁ . V₁ = P₂. V₂
1.23 atm . 35 mL = 3 atm . V₂
(1.23 atm . 35 mL / 3 atm) = V₂
V₂ = 14.35 mL
Answer:
+523 kJ.
Explanation:
The following data will be used to calculate the average C-S bond energy in CS2(l).
S(s) ---> S(g)
ΔH = 223 kJ/mol
C(s) ---> C(g)
ΔH = 715 kJ/mol
Enthalpy of formation of CS2(l)
ΔH = 88 kJ/mol
CS2(l) ---> CS2(g)
ΔH = 27 kJ/mol
CS2(g) --> C(g) + 2S(g)
So we must construct it stepwise.
1: C(s) ---> C(g) ΔH = 715 kJ
2: 2S(s) ---> 2S(g) ΔH = 446 kJ
adding 1 + 2 = 3
ΔH = 715 + 446
= 1161 kJ
3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
adding (reversed 3) from 4 = 5
ΔH = -1161 + 88
= -1073 kJ
5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
6: CS2(l) ---> CS2(g) ΔH = 27 kJ
adding 5 + 6 = 7
ΔH = -1073 + 27
= -1046 kJ
7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ
Reverse and divide by 2 for C-S bond enthalpy
= -(-1046)/2
= +523 kJ.
In this question we are asked to round off the figure 0.02056 to three significant figures.
We are given with four options to chose best out of four
a) 0.0205
b) 0.0206
c) 0.021
d) 0.020
In this question, option (b) is correct i.e. 0.0206.
There are four significant rules, we should keep in mind during round off:
- Non-zero digits are so significant
- Zeros between two non-zero digits are significant
- Leading zeros are not significant
- Trailing zeros are important only if the number contains a decimal point
To know more about Round off:
brainly.com/question/24234983
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Answer:I think that the exothermic things are vaporization and melting and the endothermic things are freezing, sublimation, condensation.
Explanation: Why? Because two of them have to do with hot and the other three have to do with cold
Answer:How many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al(s) + 3FeO(s) = 3Fe(s) + Al2O3(s). 0.8 mols.
Explanation:
