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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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The electron releases energy.
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3 years ago
The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC an
ser-zykov [4K]

Answer:

24540\frac{mg}{m^3}

Explanation:

Hello,

In this case, since the 3% by volume is represented as:

\frac{3L\ CO}{L\ gas}

By using the ideal gas equation we compute the density of CO:

\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L

Then we apply the conversion factors as follows:

=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}

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4 0
3 years ago
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krok68 [10]

Answer:

D group

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3 0
3 years ago
In plant cells, how are chloroplasts and mitochondria related?
m_a_m_a [10]

Answer:

They are very similar because they both are very important for the sustenance of the cell. Both also help to provide energy for the cell, Chloroplasts can make its own food through photosynthesis and the mitochondria helps to break down the energy.

Explanation:

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3 years ago
Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of
DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization  = 35.2 kJ/mol

Changing to  J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm)  = 4,233.822468  x-0.0005466866

ln (P2/ 1 atm)=  -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm  x760 = 75.09 torr

7 0
3 years ago
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