Calculate the number of oxygen atoms in 2.00 moles of Mn2O7

1 answer:

7 0

First you should know that there is seven oxygen atoms in one Mn2O7
So
2.00 moles of Mn2O7 contain 14.00 moles of oxygen...
Then you multiply this no. with Avagadro no....
from formula
Number of moles= no. of particles/avagadro's no..
14.00×6.02×10²³=84.28 atoms of oxygen...

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Answer:

There are 29.4 grams of oxygen in the container

Explanation:

Step 1: Data given

Volume = 20.0 L

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Temperature = 22.0 °C

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Step 2: Ideal gas law

p*V = n*R*T

⇒ p = the pressure of the gas = 845 mmHg = 1.11184

⇒ V = the volume of the gas = 20.0 L

⇒ n = the number of moles = TO BE DETERMINED

⇒R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C + 273 = 295 Kelvin

n = (p*V)/(R*T)

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n = 0.9186 moles

Step 3: Calculate mass of NO2

Mass of O2 = Moles O2 * Molar mass O2

Mass of O2 = 0.9186 moles * 32 g/mol

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A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/

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Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

2.67%

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

You are calculating weight /Volume not weight/weight % or percent by mass of the solute

Here the weight/weight % or percent by mass of the solute is asked : So first convert the VOLUME OF SOLUTION into MASS

Density of potassium nitrate in water KNO3 = 2.11 g/mL

$density=\frac{mass}{volume}$