To get the molarity, you divide the moles of solute by the litres of solution.
Molarity
=
moles of solute
litres of solution
For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.
To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.
To calculate molarity:
Calculate the number of moles of solute present.
Calculate the number of litres of solution present.
Divide the number of moles of solute by the number of litres of solution.
Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
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Answer:</h3>
1.9 moles
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Explanation:</h3>
Carbon dioxide (CO₂) is a compound that is made up of carbon and oxygen elements.
It contains 2 moles of oxygen atoms and 1 mole of carbon atoms
Therefore;
We would say, 1 mole of CO₂ → 2 moles of Oxygen atoms + 1 mole of carbon atoms
Thus;
If a sample of CO₂ contains 3.8 moles of oxygen atoms we could use mole ratio to determine the moles of CO₂
Mole ratio of CO₂ to Oxygen is 1 : 2
Therefore;
Moles of CO₂ = 3.8 moles ÷ 2
= 1.9 moles
Hence, the moles of CO₂ present in a sample that would produce 3.8 moles of Oxygen atoms is 1.9 moles
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
