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zavuch27 [327]
2 years ago
12

Determine the oxidation state of Cl in CIO3. A +3 B +7 C 0 D +1 E +5

Chemistry
1 answer:
Crazy boy [7]2 years ago
4 0

Answer:

E.) +5

Explanation:

Oxygen always has -2 oxidation number.

Because there are 3 oxygen atoms present, this means oxygen is contributing a -6 charge (-2 x 3 = -6).

Therefore, since the overall molecule is -1, chlorine must have an oxidation number of +5 to cancel all of the negative charges but 1.

You can also think of the problem like an equation. In this equation, "x" is the oxidation number of chlorine, (-2) is the oxidation number of oxygen, (3) is the number of oxygen atoms present, and the equation is set equal to (-1) because that is the overall charge of the molecule.

x - 2(3) = -1

x - 6 = -1

x = 5

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10g of hyrogen react with excess of oxygen gas according to the equation:
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Answer:

Volume of O₂ = 56 dm³

mass of water vapors (H₂O) = 90 g

Explanation:

Data Given:

mass of Oxygen = 10 g

Volume of Oxygen = ?

mass of the water vapor = ?

Reaction Given:

                 2H₂+O₂---->2H₂O

Solution:

First we have to look at the reaction for the information required

                 2H₂   +  O₂  -------> 2H₂O

               2 mol    1mol           2 mol

now convert moles to grams

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of O₂ = 2(16) = 32 g/mol

molar mass of H₂0 = 2(1) + 16 = 18 g/mol

So the masses will be

                      2H₂          +              O₂        ------->      2H₂O

                2 mol (2 g/mol)      1mol (32 g/mol)         2 mol (18 g/mol)

                      4 g                            32 g                           36 g

So now we know that

4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.

By using above information

First we find the volume of Oxygen:

For this first we find mass and then moles of Oxygen

As we know

if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen

Apply unity formula

                        4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂

                         10 g of hydrogen H₂ ≅ X g of Oxygen O₂    

by doing Cross multiplication

                         g of Oxygen O₂   = 32 g x 10 g / 4 g

                         g of Oxygen O₂   = 80 g

So,                  

mass of oxygen = 80 g

now find moles of oxygen

formula used:

            no. of moles = mass in grams/ molar mass . . . . . . (1)

Put values in above equation 1

            no. of moles = 80 g / 32 g/mol

             no. of moles = 2.5

Now to find volume of oxygen

Formula used

 Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)

Put values in equation 2

             Volume of O₂ = 2.5 moles x 22.4 dm³/mol

            Volume of O₂ = 56 dm³

______________________

Now to find mass of water vapors

As we now

if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen

Apply unity formula

                        4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)

                         10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)  

by doing Cross multiplication

                         g of water vapors (H₂O) = 36 g x 10 g / 4 g

                         g of water vapors (H₂O)   = 90 g

So,                  

mass of water vapors (H₂O) = 90 g

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