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Oduvanchick [21]
3 years ago
10

Which of the following answers below is the correctly balanced half-reaction for the reduction of the SO4 ^-2 ion in acid soluti

on? The answer is C, but I don’t understand why. Thanks in advance :-)

Chemistry
1 answer:
Rom4ik [11]3 years ago
8 0

Answer:

C

Explanation:

The oxidation number of Sulphur in SO4^2- is;

x + 4(-2) = -2

x - 8 = -2

x = -2 + 8

x = 6

Now,

the oxidation number of sulphur in H2SO3 is

2 (1) + x + 3(-2) = 0

2 + x -6 = 0

-4 + x = 0

x = 4

Hence, the oxidation number of sulphur changed from +6 to +4 which signifies  gain of two electrons as shown in option C.

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A flexible container at an initial volume of 6.13 L contains 6.51 mol of gas. More gas is then added to the container until it r
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Answer:

the final mole of the flexible container = 12.92 moles

Explanation:

Given that :

initial volume of a flexible container = 6.13 L

initial mole of a flexible container = 6.51 mol

final volume of a flexible container = 18.3 L

final mole of a flexible container = ???

Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Therefore,

n= \dfrac{V_2*n_1}{V_1}

n= \dfrac{18.3*6.51}{6.13}

n = 19.43

n=n_1+n_2

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n₂ = 19.43 - 6.51

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Answer:

Yes

Explanation:

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Answer:

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Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

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