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Murljashka [212]
3 years ago
10

Convert 4.50 e23 atoms of CO to grams.

Chemistry
1 answer:
hichkok12 [17]3 years ago
7 0

Answer:

10.4664 grams of CO

Explanation:

Remark

There's a couple of things you must look out for in this question.

1. The use of the term atoms. There are 2 atoms in each mol of CO

2. You need to divide by 2 to find the number of molecules which will lead to moles.

<u>Step one</u>

Divide the number of atoms by 2

4.50 e^23 / 2 atoms = 2.25 * 10^23 molecules.

<u>Step Two</u>

Find the number of moles of CO

1 mol of anything is 6.02 * 10^23 molecules in this case

x  = 2.25 * 10^23 molecules

1/x = 6.02 * 10^23/2.25 * 10 ^23            Cross multiply

1 * 2.25 * 10^23 = 6.02*10^23 * x           Divide by 6.02 * 10^23

2.25 * 10 ^ 23 / 6.02 * 10^23 = x

x = .3738 moles of CO

<u>Step Three</u>

Find the gram molecular mass of CO

C = 12

O = 16

1 mole = 12 + 16  = 28 grams.

<u>Step Four</u>

Find the number of gram in 0.3738 mols

1 mol = 28 grams

0.3738 mol = x                Cross multiply

x = 28 * 0.3738

x = 10.4664 grams

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If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct
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The correct option is: bent 109°

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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
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Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

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3 years ago
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