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3 years ago

What is the oxidation number of chromium in Cr2O7 4-?

Chemistry

2 answers:

Softa [21]3 years ago

8 0

<u>Answer:</u> The oxidation state of chromium in the given compound is +5.

<u>Explanation:</u>

Oxidation state is defined as the number which is given to an atom when it looses or gains electron. It is written as a superscript.

If the element gains electron, it will attain a negative oxidation state and if the element looses electrons, it will attain a positive oxidation state.

We are given a chemical compound having chemical formula of $Cr_2O_7^{4-}$

We take the oxidation state of chromium atom be 'x'.

Oxidation state of oxygen atom = -2

Overall charge on chemical compound = -4

Evaluating the oxidation state of chlorine atom:

$2(x)+7(-2)=-4\\\\x=+5$

Hence, the oxidation state of chromium in the given compound is +5.

nignag [31]3 years ago

3 0

Answer:

C. +5

Explanation:

Oxygen is always -2. And the sum of the oxidation numbers must be equal to the charge. So:

2x + 7(-2) = -4

2x - 14 = -4

2x = 10

x=5

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A 3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

andrew-mc [135]

Answer: The molecular formula for the given organic compound X is $C_6H_{8}O_7$

Explanation:

We are given:

Mass of $CO_2=4.13g$

Mass of $H_2O=1.13g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, = $\frac{12}{44}\times 4.13=1.13g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, $\frac{2}{18}\times 1.13=0.125g$ of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles$

Step 2: 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

For Carbon = $\frac{0.094}{0.094}=1$

For Hydrogen = $\frac{0.125}{0.094}=1.33$

For Oxygen = $\frac{0.109}{0.094}=1.16$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is $C_6H_8O_7$

Empirical mass = $6\times 12+8\times 1+7\times 16=192g$

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

Putting values in above equation, we get:

$n=\frac{192g/mol}{192g/mol}=1$

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

$C_6H_8O_7\times 1=C_6H_{8}O_7$

Thus molecular formula for the given organic compound X is $C_6H_{8}O_7$

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