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Oksi-84 [34.3K]
3 years ago
12

If a balloon with a volume of 10L at 500kpa is heated until the balloon has a temperature of 300°C, 750atm and now has a volume

of 100mL. What was the original temperature of the balloon (in °C)
Chemistry
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

103.6°C

Explanation:

We solve this problem, with the Ideal Gases Law. We know that moles of gas are not changed through time.

P . V = n . R . T

So n (number of moles) and R (Ideal Gases constant) are the same, they are cancelled. In the two different states, we can propose:

P₁ . V₁ / T₁ = P₂ . V₂ /T₂

We need to make some conversions:

100 mL . 1 L/ 1000mL = 0.1L

500 kpa . 1atm / 101.3 kPa = 4.93 atm

300°C + 273 = 573 K

We replace data:

4.93 atm . 10L / T₂ = 750 atm . 0.1L / 573K

4.93 atm . 10L = (750 atm . 0.1L / 573K) . T₂

(4.93 atm . 10L) / (750 atm . 0.1L / 573K) → T₂

T₂ = 376.65 K

We convert K to °C →  376.65 K - 273 = 103.6 °C

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Dmitriy789 [7]

Answer:

2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2

Explanation:

1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.

2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.

so now there is 3H2SO4

3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2

4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.

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Explanation:

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When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plo
dezoksy [38]

Complete question is;

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have

a. a pH of 7 at the equivalence point.

b. two equivalence points below 7.

c. no equivalence point.

d. one equivalence point.

e. two distinct equivalence points

Answer:

Option E - Two Distinct Equivalence points

Explanation:

I've attached a sample diprotic acid titration curve.

In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.

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