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3 years ago

If anyone could please help me i'd greatly appreciate it! I have tried over and over again and can not figure this out.

Chemistry

1 answer:

Papessa [141]3 years ago

7 0

Answer:

%Ionization = 1.63%

Explanation:

Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.

H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶

So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.

HONHNHOH => HONHNH⁺ + OH⁻

C(i) => 0.490M 0M 0M

ΔC => -x +x +x

C(eq) => 0.490 - x x x

≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100

Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]

1.3 x 10⁻⁶ = x²/0.490

=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴

=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%

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Answer:

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Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

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PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

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With this, we can day that Volume 2 decreases.

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