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Gala2k [10]
3 years ago
11

Pls help! 2CO + O2 → 2CO2

Chemistry
2 answers:
jok3333 [9.3K]3 years ago
7 0

The bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Since the chemical reaction is 2CO + O₂ → 2CO₂ and the total bond energy of the products carbon dioxide CO₂ is 1,472 kJ.

Since from the chemical reaction, we have 2 moles of CO₂ which gives 1,472 kJ and there are two carbon-oxygen, C-O bonds in CO₂, then

2 × C-O bond = 1,472 kJ

1 C-O bond = 1.472 kJ/2

C-O bond = 736 kJ

So, the bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Learn more about bond energy here:

brainly.com/question/21670527

olganol [36]3 years ago
7 0

Answer: 736 kJ

Explanation:

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Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

1 Tm = 10^{21} nm

4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

6 0
3 years ago
Balanced equation for solid magnesium oxide
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Answer:

Mg + O2 → MgO

Explanation:

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At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO
erastovalidia [21]

Answer : The concentration of NO at equilibrium is 0.9332 M

Solution :  Given,

Concentration of N_2 and O_2 at equilibrium = 0.200 M

Concentration of N_2 and O_2 at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}

K_c=6.25

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

                         N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially             0.200   0.200         0

.800

At equilibrium  (0.200-x) (0.200-x)  (0.800+2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}

By solving the term x, we get

x=2.6\text{ and }-0.0666

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of NO at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of NO at equilibrium is 0.9332 M

5 0
3 years ago
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