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11 months ago

A gas with a pressure of 2.25 atm occupies 450.0 mL at a temperature of 300 K. What is the volume at 405.0 K?

1 answer:

6 0

The volume of the gas at a temperature of 405.0 K would be 607.5 mL. Making option D the right answer to the question.

What is the volume of the gas?

To find the volume of the gas, the equation to be used would have to be combine gas law.

Combine gas law as the name suggest uses the combination of Charles law which measures Volume against temperature, and Gay-Lussac's law which measures Pressure/Temperature, and Boyle's law which measures pressure X volume where k is constant.

Using the combine law to find the volume, we have:

P₁V₁/T₁=P₂V₂/T₂

Where P₁ = initial pressure

V₁ = initial volume

T₁ = initial temperature

P₂ = final pressure

V₂ = final volume

T₂ = final temperature

P₁ = 2.25atm

V₁ = 450.0 mL

T₁ = 300 K

T₂ = 405.0 K

V₂ = ?

D) 607.5 mL

= [2.25(450)]÷300=[2.25(V₂]÷405

Making V₂ the subject

3.375=2.25 V₂ ÷ 405

V₂ = 3.375 x 405 ÷ 2.25

V₂ = 607.5 mL

In summary, a gas with an initial pressure of 2.25atm, an initial pressure of 450.0 mL and an initial temperature of 300 K would have a final volume of 607.5 mL if the temperature is increased to 405.0 K.

Learn more about Combine gas law here: brainly.com/question/13538773

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If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?

Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

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Given Data

Volume of Hexane dispensed = 40ml

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2 years ago

Write a balanced equation formed when the following elements react with oxygen:a)Zinc

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Explanation:

a) when zinc burnt in oxygen.

2Zn + O2 -----∆-----> 2ZnO(black residue)

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C+O2----∆---> CO2.

c) when sulphur burnt in oxygen.

S+O2-----∆-----> SO2.

d) when Calcium burnt in oxygen.

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2 years ago

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.

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So the cyclist went 800+500+1200 m = 2500 m for total distance.

10t =800 leads to t=80 sec

5t=500 leads to t=100 sec

13t = 1200 leads to t = 92.3 sec

total time is 272.3 sec

Average speed for the entire trip is 2500 / 272.3 = 9.18 m/s

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