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Nonamiya [84]
2 years ago
9

25 L of a gas is collected at 115 kPa. If the pressure increases to 300 kPa, what is the new volume?

Chemistry
1 answer:
Lina20 [59]2 years ago
5 0

Answer:

V2 = 9.58 Litres.

Explanation:

Given the following data;

Initial volume = 25 L

Initial pressure = 115 kPa

Final pressure = 300 kPa

To find the new volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

115 * 25 = 300*V_{2}

2875 = 300*V_{2}

V_{2} = \frac {2875}{300}

V_{2} = 9.58

V2 = 9.58 L

Therefore, the new volume is 9.58 litres.

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LDL is

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high density lipid

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HDL helps rid your body is excess cholesterol so so it won't end up in your arteries

LDL is also called "bad cholesterol" because it takes cholesterol to your arteries

6 0
3 years ago
Which of these does not cause a change in the reaction rate?
VMariaS [17]
A. density
<span>Reaction rate increases with concentration.</span>
8 0
2 years ago
Read 2 more answers
What mass of potassium chloride, KCl, is produced when 12.6 g of oxygen, 02, is produced?
blagie [28]

Mass of KCl= 19.57 g

<h3>Further explanation</h3>

Given

12.6 g of Oxygen

Required

mass of KCl

Solution

Reaction

2KClO3 ⇒ 2KCl + 3O2

mol O2 :

= mass : MW

= 12.6 : 32 g/mol

= 0.39375

From the equation, mol KCl :

= 2/3 x mol O2

= 2/3 x 0.39375

=0.2625

Mass KCl :

= mol x MW

= 0.2625 x 74,5513 g/mol

= 19.57 g

4 0
2 years ago
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
An acid with the general formula RCOOH is used to make a 0.10 M solution in water. The pH of this acid solution is measured as 3
Sati [7]

Answer:

0.159 \%

Explanation:

The acid will dissociate according to the reaction shown below:-

RCOOH + H_2O\rightleftharpoons RCOO^- + H_3O^+

Given that, pH=3.8

The concentration of can be determined from the expression fo pH as:-

pH = - log [H^+]

3.8  = - log [H^+]

[H^+] = 1.59\times 10^{-4}\ M

The initial concentration of RCOOH was 0.10 M, then the percent dissociation was- calculated as shown below:-

\% \ dissociation=\frac{1.59\times 10^{-4}}{0.10}\times 100=0.159\ \%

3 0
2 years ago
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