5.6 dm3 is divided by 22.4 dm3 to give number of moles:
=5.6dm3/22.4moles per dm3
=0.25 moles
So 0.25 moles of C3H4 react.
For this you need the atomic (molecular) mass of C3H4. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. C3H4=40.0 grams10.0 grams C3H4 / 40.0 grams = .250 moles C3H4
Propyne is C3H4, and combustion means burning in Oxygen, or O2 So, start with C3H4 + O2 --> ? ? Combustion results in H2O and CO2 So write as C3H4 + O2 --> H2O + CO2 There are three Carbons on the left, so make the right have three, and since there is O2 on the left, it will always be an even number of O, so try H2O, who has an odd number of O, as 2 (or any even numbers) Continue doing this until the left and right sides of the equations have the same number of C's, O's, and H's. Answer: C3H4 + 4O --> 2H2O + 3CO2
C3h4
C3h4
citric acid is C6H8O7
30 moles
For this you need the atomic (molecular) mass of C3H4. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. C3H4=40.0 grams10.0 grams C3H4 / 40.0 grams = .250 moles C3H4
Propyne is C3H4, and combustion means burning in Oxygen, or O2 So, start with C3H4 + O2 --> ? ? Combustion results in H2O and CO2 So write as C3H4 + O2 --> H2O + CO2 There are three Carbons on the left, so make the right have three, and since there is O2 on the left, it will always be an even number of O, so try H2O, who has an odd number of O, as 2 (or any even numbers) Continue doing this until the left and right sides of the equations have the same number of C's, O's, and H's. Answer: C3H4 + 4O --> 2H2O + 3CO2
It is Propyne , C3H4
C3h4
C3h4
citric acid is C6H8O7
C4H10
H2C = C = CH2 is C3H4 and it has 2 C-carbon double bonds.
1.51 x 10 ^23 molecules
C9h12
Cyclopropene, a cyclic alkeneMethylacetylene, a common alkynePropadiene, an allene