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?
1 mol CH4 x (16 g CH4/1 mole CH4) = ?g CH4
L CH4 x (1 mole CH4/22.4 L) = ? moles CH4
#molecules CH4 x (1 mole/6.022 x 10^23 molecules) = ? moles CH4
The fraction (ratio) in parentheses is the ratio the problem is asking for. You did b right. Note that in my solutions for
a). mole cancels to leave g as the unit.
b). L cancels to leave moles as the unit.
c). # molecules cancel to leave moles as the unit.
I don't kno either bro
car
well if Ch4 is is being transferred then you need to.......
idk what im saying lol
To carry out each of the conversions, you need to apply the appropriate conversion factors. Let's go through each conversion:
a. mol CH4 to grams CH4:
To convert from moles (mol) to grams (g) of a substance, you need to use its molar mass. For CH4, the molar mass is 16 g/mol. So the correct conversion factor is:
1 mol CH4 / 16 g CH4
b. L CH4(g) to mol CH4(g) at STP (standard temperature and pressure):
At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters. So the correct conversion factor is:
1 mol CH4 / 22.4 L
c. molecules CH4 to mol CH4:
Avogadro's number tells us that there are 6.02 x 10^23 molecules in 1 mole of a substance. Thus, the correct conversion factor is:
1 mol CH4 / 6.02 x 10^23 molecules CH4
Based on your solutions, it appears that you have the correct conversion factors for each conversion. Your conversion factors for a. and b. are correct, and for c., it should be 6.02 x 10^23 molecules CH4 / 1 mol CH4. So your conversions are indeed correct!