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A sample of gas occupies 10.0 L at 240°C under a pressure of

NISA [10]

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

8 0

3 years ago

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Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

2 years ago

What happens when an island volcano erupts and molten lava flows out?

qwelly [4]

It will explode together cause danger

3 0

3 years ago

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A boy weighs 525 N. What is his mass?

MAXImum [283]

Answer:≈40.8 kg.

Explanation:

4 0

3 years ago

How many grams of agcl would be needed to make a 4. 0 m solution with a volume of 0. 75 l?

devlian [24]

430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

<h3>What is Molarity?</h3>

The amount of a substance in a specific volume of solution is known as its molarity (M).

The number of moles of a solute per liter of a solution is known as molarity.

<h3>Calculation of Required amount of AgCl</h3>

Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

0.75L×4.0M=3.0mol

Then, using the periodic table's molar mass for AgCl, convert from moles to grams:

3.0mol×143.321gmol=429.963g

The final step is to round to the correct significant figure, which in this case is two: 430g.

Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

Learn more about Molarity here:
brainly.com/question/8732513

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8 0

1 year ago