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timurjin [86]
2 years ago
15

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

er a large quantity of heated Copper (II) oxide, CuO, the compound A is completely oxidized to carbon dioxide and water. Copper is the only other product of the reaction. The products are collected and it is found that 0.352g of CO2 and 0.144g of H2O are formed. Questions :
1) Calculate the mass of carbon present in 0.352g of CO2
2) Use this value to calculate the amount in moles of carbon atoms present in 0.240g of A
Chemistry
1 answer:
vesna_86 [32]2 years ago
4 0

(a)   In this section, give your answers to three decimal places.

(i)

Calculate the mass of carbon present in 0.352 g of CO

2

.

Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g

of

A

.

(ii)

Calculate the mass of hydrogen present in 0.144 g of H

2

O.

Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g

of

A

.

(iii)

Use your answers to calculate the mass of oxygen present in 0.240 g of

A

Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g

of

A


(b)

Use your answers to

(a)

to calculate the empirical formula of

A


thank you

hope it helpsss

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According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams
gogolik [260]

Answer:

m_{HClO_3}=12.7gHClO_3

Explanation:

Hello,

Considering the reaction:

3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3

The molar masses of chlorine and chloric acid are:

M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3

Best regards.

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2 years ago
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NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol×\frac{1mol}{80,043g}X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0
2 years ago
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