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What ratio is used to carry out each conversion? a.mol CH4 to grams CH4 b.L CH4(g) to mol CH4(g)(at STP) c.molecules CH4 to mol CH4 I honestly don't kno what it's asking me I got 1 mol CH4/ 16(g) CH4 for a. 1 mol/22.4 L for b. and 6.02 x 10^23 molecules/1 mol but I don't think this is right, is it?

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MissPhiladelphia
a.mol CH4 to grams CH4
To convert from moles to grams, we need the molar mass of CH4 which is 16.05 g/mol.


b.L CH4(g) to mol CH4(g)(at STP)
To convert from L to mol CH4, we need the relation 1 mol of an ideal gas is equal to 22.4 L of that gas. The ratio would be 1/22.4 mol/L.


c.molecules CH4 to mol CH4
To convert from molecules to mol, we need the avogadro's number which is 6.022x10^23 units / 1 mol. The ratio to be multiplied would be 1/6.022x10^23 mol/molecules.

Explanation:

According to mole concept

a. [tex]CH_4[/tex]Moles to [tex]CH_4[/tex]grams

Number of moles =[tex]\frac{\text{Mass of the compound}}{\text{Molar mass of the compound}}[/tex]

Mass of [tex]CH_4[/tex]= Moles of [tex]CH_4\times [/tex] 16 g/mol

b.[tex]CH_4[/tex] Liters to [tex]CH_4[/tex] moles at STP.

At STP,1 mol gas occupies 22.4 L of volume.

Multiply the given volume of gas in L with [tex]\frac{1 }{22.4 } moles[/tex]

c.Molecules [tex]CH_4[/tex]of to moles [tex]CH_4[/tex]

1 mole = [tex]N_A=6.022\times 10^{23} [/tex] atoms or molecules

So, in 1 molecule =[tex]\frac{1}{N_A}=\frac{1}{6.022\times 10^{23}} [/tex] mole.

Multiply the given number of molecules with [tex]\frac{1}{6.022\times 10^{23}} [/tex] mole to get the number of moles of [tex]CH_4[/tex].

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