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Arte-miy333 [17]
2 years ago
7

A gas has a volume of 50.0 cm3 at a temperature of -73°c. what volume would the gas occupy at a temperature of -123°c if the pre

ssure stays constant?
Chemistry
1 answer:
vaieri [72.5K]2 years ago
7 0

Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula  p1V1 = P2V2 , where P1V1 denotes initial pressure and volume  and P2V2 denotes values of pressure and  volume.

Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p 
V = 50.0 p / 2p

= 50.0 /2

= 25.0 m^3 
b. if the pressure is cut in half
50.0 p = V x p/2 
100 p = V x p 
V = 100 m^3 
c. if the pressure is tripled
50.0 p = V x 3p 
V = 50.0 p / 3p

= 50.0 /3

=16.7 m^3 

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3 years ago
How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m
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<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})  

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=?mol  

[CH_3COOH]=0.200mol

pH = 5.00

Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

[CH_3COONa]=0.364mol

To calculate the mass of sodium acetate for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

Hence, the mass of sodium acetate that must be added is 30.23 grams

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3 years ago
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Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C

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2 years ago
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Answer:

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Explanation:

This is the reaction:

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