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QUESTION 22 What is the oxidation number of Cl in HClO 2 ? +5 +3 -2 +4 -4 0.7143

ID: 833442 • Letter: Q

Question

QUESTION 22

What is the oxidation number of Cl in HClO2?

+5

+3

-2

+4

-4

0.7143 points   

QUESTION 23

What period 3 element having the following ionization energies (all in kJ/mol)?

IE1 = 1012;   IE2 = 1900; IE3 = 2910; IE4= 4960; IE5= 6270; IE6 = 22,200

P

Mg

S

Si

Cl

0.7143 points   

QUESTION 24

Write the formula for copper(II) sulfate heptahydrate.

Cu2SO . H2O

Cu2SO3. H7

CuSO4.7H2O

CuS . 5H2O

(CuSO4)II

0.7143 points   

QUESTION 25

What mass of oxygen (in kilograms) is contained in 3.00 gallons of ethanol (C2H5OH)? (The density of ethanol is 0.789 g/mL; 1 gal = 3.785 L).

8.50 x 10-3 kg O

9.97 x 103 kg O

2.03 kg O

3.11 kg O

1.94 kg O

0.7143 points   

QUESTION 26

Which one of the following compounds is insoluble in water?

NaCl

KI

KNO3

AgCl

No correct answer is given.

0.7143 points   

QUESTION 27

How many different values of l are possible in the third principal level?

4

3

0

2

1

0.7143 points   

QUESTION 28

According to the following reaction, how much energy is evolved during the reaction of 32.5g B2H6 and 72.5 g Cl2?  [The molar mass of B2H6 is 27.67 g/mol. The molar mass of Cl2 is 70.90 g/mol.]

             B2H6(g)   +   6 Cl2(g)  =>     2 BCl3(g) + 6 HCl(g)        ?Horxn= -1396 kJ

238 kJ

3070 kJ

1640 kJ

429 kJ

1430 kJ

+5

+3

-2

+4

-4

Explanation / Answer

1. 3

The oxidation state of O is always -2 except when in peroxides and F2O.
Since there are 2 O, the total oxidation states for O is 2 x (-2) = -4.

The overall charge for HClO2 is 0. Therefore, the total sum of all the oxidation states is equal to 0. Hence,
(+1) + (Cl) + (-4) = 0
(Cl) + (-3) = 0
(Cl) = +3

2. Phosphorus as it is increasing.

3.CuSO4.7H2O

4. 3 gallons = 11.355 litres

so mass 0.789 x 1000x 11.355 = 8959 grams of ethanol

ethanol mass is 46 and oxygen mass is 16

so it will be 16*8959/46 = 3116 grams or 3.11 kg

5. agcl generally all compounds of silver do not dissolve in water or have limited solubility in it

6. 3 (0,1,2)

7.cl2 is limiting agent here.

so it will be 1396*72.5/70.9*6 = 238 kJ

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