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2 years ago

3. The law of definite proportions helps us explain why water will always

2 answers:

3 0

C.) The law of definite proportions helps us explain why water will always "<span>consist of a particular proportion of hydrogen and oxygen."

Hope this helps!</span>

Vesna [10]2 years ago

3 0

Um I Think its C I guessed on the test and I got it right and i think thats the one i picked

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Why does increasing the number of trials increase confidence in the results of the experiment?

valkas [14]

<span>It rises confidence for the reason that the more times you conduct the similar experiment over and over should either demonstrate your hypothesis right and wrong and remove any random incidences that might touch your results. Meaning it permits to have a more accurate measure or conclusion.</span>

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2 years ago

Read 2 more answers

15. An apparatus consists of a 4.0 dm3

Solnce55 [7]

Answer:

i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·

ii· Total pressure of the gas mixture is 253.143kPa·

Explanation:

<h3>solution for i :</h3>

Assuming that the given gases to be ideal,

so by ideal gas equation

PV=nRT

where,

P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume

As the temperature of both gases are same, the mixing process is an isothermal process

PV=constant

Initially for nitrogen gas PV=803×4=3212

let P $x_{1}$ be the partial pressure of the nitrogen gas

(P $x_{1}$ )×14=3212

∴P $x_{1}$ =219.429kPa

∴Partial pressure of nitrogen gas is 219.429kPa

Initially for argon gas PV=47.2×10=472

let P $x_{2}$ be the partial pressure of the argon gas

(P $x_{2}$ )×14=472

∴P $x_{2}$ =33.714kPa

∴Partial pressure of argon gas is 33.714kPa

solution for ii :

Total pressure of the gas mixture will be the sum of the partial pressures of each gas as

Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)

∴Total pressure=P $x_{1}$ +P $x_{2}$

=219.429+33.719

Total pressure =253.143kPa

6 0

2 years ago

Read 2 more answers

Which models of the atom include a structure that is mostly made of empty space?

irina1246 [14]

Thomson proposed this idea. Rutherford disproved it when he shot alpha particles at a gold sheet and they bounced back at him, proving that the alpha particles bounced off something aka the nucleus.

D.

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2 years ago

1. Calculate the molarity of a sugar solution if 4 liters of the solution contains 8 moles of sugar

klio [65]

Answer:

The molarity of a sugar solution is 2 M.

Explanation:

Molarity is a concentration measure that expresses the moles of solute per liter of solution. In this case it is calculated with the simple rule of three:

4 L of solution--------8 moles of sugar

1 L of solution ------x= (1 L of solution x 8 moles of sugar)/4 L of solution

x=2 moles of sugar---> <em>The solution is 2M</em>

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2 years ago

A reaction container holds 5.77 g of P4 and 5.77 g of O2.

Dvinal [7]

Answer:

a) O2 is the limiting reactant

b) 5.75 grams P4O10

c) 5.79 grams P4O6

Explanation:

Step 1: Data given

Mass of P4 = 5.77 grams

Mass of O2 = 5.77 grams

Molar mass of P4 = 123.90 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

P4 + 3O2 → P4O6

Step 3: Calculate moles of P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 5.77 grams / 123.90 g/mol

Moles P4 = 0.0466 moles

Step 4: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 5.77 grams / 32.0 g/mol

Moles O2 = 0.1803 moles

Step 5: Calculate limiting reactant

P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

There will remain 0.1803 - 0.1398 = 0.0405 moles O2

Step 6: Calculate the amount of P4O6

For 1 mol P4 we'll have 1 mol P4O6

For 0.0466 moles P4 we'll have 0.0466 moles P4O6

Step 7: The balanced equatio

P4O6 + 2O2 → P4O10

We have 0.0466 moles P4O6 and 0.0405 moles O2

Step 8: Calculate the limiting reactant

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

Step 9: Calculate moles and mass of P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

For 0.0405 moles O2 we'll have 0.02025 moles P4O10

This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10

3 0

2 years ago