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3 years ago

A 4.75 Liter container is at a pressure of 120 kPa. What will its pressure be at a volume of 1.85L?

Chemistry

1 answer:

AlladinOne [14]3 years ago

8 0

Answer:

The answer to your question is 308.1 kPa

Explanation:

Data

Volume 1 = V1 = 4.75 l

Pressure 1 = P1 = 120 kPa

Volume 2 = V2 = 1.85 l

Pressure 2 = ?

Process

To solve this problem use Boyle's law.

P1V1 = P2V2

- Solve for P2

P2 = P1V1 / V2

- Substitution

P2 = (120 x 4.75) / 1.85

-Simplification

P2 = 570 / 1.85

-Result

P2 = 308.1 kPa

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620 mL of nitrogen at standard pressure is compressed into a 480 mL container. What is the new pressure in kPa?

bulgar [2K]

Answer:

c. 131 kPa

Explanation:

Hello!

In this case, since the relationship between volume and pressure is inversely proportional, based on the Boyle's law:

$P_1V_1=P_2V_2$

Considering that the standard pressure is 101.325 kPa, we can compute the final pressure as shown below:

$P_2=\frac{P_1V_1}{V_2}=\frac{620mL*101.325kPa}{480mL}\\\\P_2=131kPa$

Therefore, the answer is c. 131 kPa .

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2 years ago

If a student did not remove all of the bubbles from inside the buret before reading the initial volume and beginning the titrati

matrenka [14]

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If there's an air bubble inside the buret, and the bubble escapes the buret during the titration the initial volume lecture (Vi) would be lower (closer to 0) than the actual one, and the recorded consumed volume (ΔV=Vf-Vi) would be higher than the actual one and thus the calculated concentration of the hydrochloric acid would be higher than the real one.

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3 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

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3 years ago

Mi One should know the properties of the components of the mixture to separate it. Explain with an example.

Alekssandra [29.7K]

One should know that the properties of the components of the mixture.
Suppose a mixture contains Sodium or potassium .If you put them in open air it may catch fire.
Suppose a mixture has flammable components like kerosene,spirit .If you put them in fire it may harm you .

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3 years ago

At 25.5°C a gas has a volume of 125mL. What would be the volume be if the temperature increased to 50°C

kolezko [41]

Answer:

245 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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3 years ago