I am pretty sure it is d if i am wrong please tell me
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.
Answer: it measures temperature
Explanation:
Answer:
C. Ca.
Explanation:
The atom loses or gains electrons forming anion or cation to reach stability (fill the outermost shell of the electron levels).
When atom gains electrons forms anions with negative charge.
While, when atom loses electrons forms cations with positive charge.
A. O:
O atom has 8 electrons in its nucleus and gains 2 electrons to reach stability (10 electrons, configuration of Ne).
<em>So, it forms O²⁻.</em>
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B. AI:
Al atom has 13 electrons in its nucleus and loses 3 electrons to reach stability (10 electrons, configuration of Ne).
<em>So, it forms Al³⁺.</em>
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C. Ca:
Ca atom has 12 electrons in its nucleus and loses 2 electrons to reach stability (10 electrons, configuration of Ne).
<em>So, it forms Ca²⁺.</em>
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D. Na:
Na atom has 11 electrons in its nucleus and loses 1 electron to reach stability (10 electrons, configuration of Ne).
<em>So, it forms Na⁺.</em>
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