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andreyandreev [35.5K]

3 years ago

9

A mixture of 4 moles of CO2 and 10 moles of H2O are allowed to reach equilibrium in a 10. L container. What are the concentratio

ns (M) of all species at equilibrium?
2CO2 (g) + H2O (g) ← → 2O2 (g) + CH2CO (l) K = 6.1 x 10-8

1 answer:

il63 [147K]3 years ago

7 0

Answer:

I believe the answer your looking for is 12.2

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How are mitosis and meiosis similar? How are they different? Provide 1 way the two processes are the same and 2 ways the 2 proce

swat32

Answer:

Similar: both processes of cell division; both processes take place in the nucleus of the cell

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Explanation:

https://byjus.com/biology/mitosis-and-meiosis/

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3 years ago

If you have a 150-gram sample of CrO3: • How many moles of CrO3 do you have? • How many oxygen atoms do you have? • How many gra

docker41 [41]

Answer :

(a) The moles of $CrO_3$ is, 1.5 moles.

(b) The number of oxygen atoms are, $27.099\times 10^{23}$

(c) The mass of oxygen is, 72 grams.

Explanation : Given,

Mass of $CrO_3$ = 150 g

Molar mass of $CrO_3$ = 100 g/mole

Molar mass of oxygen = 16 g/mole

(a) First we have to calculate the moles of $CrO_3$ .

$\text{Moles of }CrO_3=\frac{\text{Mass of }CrO_3}{\text{Molar mass of }CrO_3}=\frac{150g}{100g/mole}=1.5moles$

The moles of $CrO_3$ is, 1.5 moles.

(b) Now we have to calculate the number of oxygen atoms.

In $CrO_3$ , there are 1 atom of chromium and 3 atoms of oxygen.

According to the mole concept,

1 mole of $CrO_3$ contains $3\times 6.022\times 10^{23}$ number of oxygen atoms.

So, 1.5 mole of $CrO_3$ contains $1.5\times 3\times 6.022\times 10^{23}=27.099\times 10^{23}$ number of oxygen atoms.

The number of oxygen atoms are, $27.099\times 10^{23}$

(c) Now we have to calculate the mass of oxygen.

As, 1 mole of $CrO_3$ contains 3 moles of oxygen

So, 1.5 mole of $CrO_3$ contains $1.5\times 3=4.5$ moles of oxygen

$\text{Mass of oxygen}=\text{Moles of oxygen}\times \text{Molar mass of oxygen}$

$\text{Mass of oxygen}=(4.5mole)\times (16g/mole)=72g$

The mass of oxygen is, 72 grams.

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