A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.0500% carbon dioxide, and 0.930% argon by volume. How many molecules of each gas are present in 6.51 L of the sample at 43°C and 1.59 atm? Enter your answers in scientific notation.

Answer : The number of molecules of nitrogen, oxygen, carbon dioxide and argon is, , , and respectively.

Explanation :

First we have to calculate the total moles of mixture of gas by using ideal gas equation.

where,

P = Pressure of mixture of gas = 1.59 atm

V = Volume of mixture of gas = 6.51 L

n = number of moles mixture of gas = ?

R = Gas constant =

T = Temperature of mixture of gas =

Putting values in above equation, we get:

Now we have to calculate the moles of nitrogen, oxygen, carbon dioxide and argon.

Moles of nitrogen =

Moles of oxygen =

Moles of carbon dioxide =

Moles of argon =

Now we have to calculate the number of molecules of nitrogen, oxygen, carbon dioxide and argon.

As, 1 mole of nitrogen contains number of molecules of nitrogen.

So, 0.312 mole of nitrogen contains number of molecules of nitrogen.

and,

As, 1 mole of oxygen contains number of molecules of oxygen.

So, 0.0836 mole of oxygen contains number of molecules of oxygen.

and,

As, 1 mole of carbon dioxide contains number of molecules of carbon dioxide.

So, 0.000199 mole of carbon dioxide contains number of molecules of carbon dioxide.

and,

As, 1 mole of argon contains number of molecules of argon.

So, 0.00371 mole of argon contains number of molecules of argon.

Therefore, the number of molecules of nitrogen, oxygen, carbon dioxide and argon is, , , and respectively.

i'm wiling to try and you : )

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l