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sergeinik [125]
3 years ago
13

When carbon is oxidized in a small amount of oxygen, the principle product formed is carbon monoxide (co). when the oxidation oc

curs in a higher concentration of oxygen, the principle product formed is carbon dioxide (co2). the mass ratio of oxygen to carbon in carbon dioxide (co2) is twice the mass ratio of oxygen to carbon in carbon monoxide (co)?
Chemistry
1 answer:
skad [1K]3 years ago
6 0
<span>It generally does not mean that there is double the oxygen, but in this case there is double, because the subscript number tells how many atoms of that element are in a particle. In this case, there are two of the oxygen, hence the DI-oxide verbiage, and one of the carbon. When there is only one, it's MONOxide, to indicate only one atom.</span>
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There is 7 moles in C
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4vir4ik [10]

Answer: 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles

The balanced chemical reaction is:

2AuCl_3\rightarrow 2Au+3Cl_2  

According to stoichiometry :

2 moles of AuCl_3 produce =  3 moles of Cl_2

Thus 0.242 moles of  will produce= \frac{3}{2}\times 0.242=0.363mol of Cl_2

Mass of Cl_2= moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g

Thus 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

5 0
2 years ago
What is the purpose of the following in the experiment of Synthesis of Methyl Salicylate
Aneli [31]

Answer:

1. Phosphoric Acid : Catalyst

2. Methyl Anthranilate : Reactive

3. Sodium Nitrite : Reactive

4. Diethyl Ether : Solvent and reactant

5. Nitrogen : Sub-product

Explanation:

The phosphoric acid is used as a catalyst for the reaction, the methyl anthranilate will react with the sodium nitrite to produce methyl salicylate, along with the diethyl ether and the nitrogen is a sub-product of the reaction.

7 0
2 years ago
A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure
ycow [4]

<u>Answer:</u> The molecular formula for the compound is C_6H_6

<u>Explanation:</u>

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = \frac{7.68}{7.68}=1

For Hydrogen = \frac{7.74}{7.68}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

  • <u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{m}{M}RT

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = 62.3637\text{ L. torr }mol^{-1}K^{-1}

T = temperature of the gas = 100^oC=(100+273)K=373K

Putting values in above equation, we get:

820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

n=\frac{78.10g/mol}{13g/mol}=6

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 6)}H_{(1\times 6)}=C_6H_6

Hence, the molecular formula for the compound is C_6H_6

8 0
3 years ago
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