Answer:
(1) -12 Kcal/mol
Explanation:
Our answer options for this question are:
(1) -12 Kcal/mol
(2) -13 Kcal/mol
(3) -15 Kcal/mol
(4) -16 Kcal/mol
With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, <u>two bonds are broken</u>, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are <u>formed</u>.
If we want to calculate the enthalpy value, we can use the equation:
<u>ΔH=ΔHbonds broken-ΔHbonds formed</u>
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
<u>ΔHbonds broken</u>
<u />
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
<u>ΔHbonds formed</u>
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
<u>ΔH of reaction</u>
<u />
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it helps!
<u />
A good night to see him again soon as I have to work tomorrow morning and I’m just gonna let go out and get it out and then I’ll head out and I can do that and then I’ll
(a) Reaction of nitric acid with non-metal:
C+4HNO
3
⟶CO
2
+2H
2
O+4NO
2
S+6HNO
3
⟶H
2
SO
4
+2H
2
O+6NO
2
(b) Nitric acid showing acidic character:
K
2
O+2HNO
3
⟶2KNO
3
+H
2
O
ZnO+2HNO
3
⟶Zn(NO
3
)
2
+H
2
O
(c) Nitric acid acting as oxidizing agent
P
4
+20HNO
3
⟶4H
3
PO
4
+4H
2
O+20NO
2
3Zn+8HNO
3
⟶3Zn(NO
3
)
2
+4H
2
O+2NO
hope that helps you please mark brainliest
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