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artcher [175]
3 years ago
10

What is the percent of O in CO2

Chemistry
2 answers:
zmey [24]3 years ago
5 0

72.71 . simple google search, you shouldnt waste points

MrMuchimi3 years ago
4 0

For this case we have the CO_ {2}

It is composed of one atom of carbon and two of oxygen.

The atomic mass of carbon and oxygen, which can be found in a periodic table, are given by:

C: 12 \frac {g} {mol}\\O: 16 \frac {g} {mol}

Then, we find the atomic mass of CO_{2}:

1 * 12 \frac {g} {mol} = 12 \frac {g} {mol}\\2 * 16 \frac {g} {mol} = 32 \frac {g} {mol}

Adding we have:

44 \frac {g} {mol}

To find the percentage of oxygen, we divide the atomic mass of the two oxygen atoms between that of CO_{2}:

\frac {32} {44} * 100 = 72.73%

Thus, the percentage of oxygen is 72.73%

Answer:

72.73%

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Then by applying water,
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In a titration of 0.5 M HCl and an unknown concentration NaOH, 25.0 ml of NaOH was required to completely neutralize 25.0 ml of
baherus [9]
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Brian is testing to see how much water plants need. Why is it important that Brian includes a control group? A) To observe wheth
bija089 [108]

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7 0
2 years ago
Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

5 0
3 years ago
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