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3 years ago

Help ASAP please!!!!!!!!!!!

Chemistry

1 answer:

iris [78.8K]3 years ago

7 0

Answer:

The second one

Explanation:

Gas pressure is caused by gas molecules bouncing off the container walls and each other. Every time a molecule changes direction because it hits a wall, the change in momentum results in a small push. Due to the large number of molecules involved, the pushes add up to a large amount of pressure.

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He partial pressure of oxygen gas in our atmosphere is 0.21 atm. this is the partial pressure at which human lungs have evolved

solong [7]

Hello!

The partial pressure of helium to keep the partial pressure of oxygen at 0,21 atm in a scuba-diver tank is 8,09 atm

To solve this question, we can use the Dalton's Law, which states that the total pressure in a container with a mixture of gases is the sum of the partial pressures o each individual gas. For the case of this mixture the Dalton's Law is as follows:

$P_{tot}=P_{He} +P_{O_2}$

In this equation, we need to clear for PHe, knowing that the PO₂ should be 0,21 atm, to find the required pressure of Helium:

$P_{He}=P_{tot} -P_{O_2}=8,30atm-0,21atm= 8,09 atm$

Have a nice day!

6 0

3 years ago

Why do we need a “mole”?

Ivan

Wdym mole like the animal or something else ?

4 0

2 years ago

Read 2 more answers

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.

4 0

3 years ago

A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

pshichka [43]

Answer: Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025 Ca(NO3)2 >> Ca2+ + 2NO3- Moles NO3- = 2 x 0.025 = 0.05 Moles HNO3 = 400 x 0.100 / 1000 = 0.04 Total moles = 0.05 + 0.04 = 0.09 Total volume = 500 ml = 0.500 L M = 0.09 / 0.500 = 0.18

6 0

4 years ago

Convert all of the following lengths to millimeters.

xenn [34]

A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm

8 0

2 years ago

Read 2 more answers