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Leviafan [203]
2 years ago
13

How many moles are in 8.32 × × times 10^24 molecules of co2?

Chemistry
1 answer:
Gnom [1K]2 years ago
7 0

Answer:

13.8 moles

Explanation:

You know that in a mole you have 6.022x10²³ something in this case of molecules. By knowing that by each mole you have 6.022x10²³ molecules, then you can establish the factor of conversion:

8.32x10²⁴molec of CO2 x 1 mol/6.022x10²³ molec of CO2

You divide 8.32x10²⁴/6.022x10²³ and you get 13.81 moles.

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Volcanic activity during Earth’s formation is responsible for contributing a significant portion of CO2 and H2O to the atmospher
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The answer is A i think if it’s wrong someone correct me
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2 years ago
A sample of 0.370 mol of a metal oxide (m2o3) weighs 55.45 g. how many grams of o are in the sample?
Wewaii [24]
0.370 mol metal oxide = 55.45 g 

<span>1 mol = 55.45/0.370 = 149.86 g </span>

<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>

<span>there is 48/149.86 * 100% O in the sample </span>

<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
6 0
3 years ago
( i need help on this one )
lara [203]

Answer:

1. False

2.True

Explanation:

7 0
3 years ago
An object has a mass of 18.4g and a volume of 11.2 ml what is the density
DENIUS [597]

Density=mass/volume

Density=18.4g/11.2ml

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Just make sure you include the unit of measurement with your answer.

4 0
3 years ago
Read 2 more answers
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

8 0
2 years ago
Read 2 more answers
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