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Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

The molar mass will be thus:

Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

The percent composition of Carbon is thus 12.7 %.
Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction: