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Answered 2017-07-20 15:58:39
It is either n-butane
or
methyl propane (i.e. 'iso-butane' )
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C4H10(g) is about -126 kJ/mol C4H10(l) is about -147 kJ/mol
C4h10 + 6.5o2 -> 4co2 + 5h2o
1 mole C4H10 = 58.1222g = 6.022 x 1023 molecules 11.7g C4H10 x 6.022 x 1023 molecules/58.1222g = 1.21 x 1023 molecules C4H10
Butane-C4H10 C4H10 - H C4H9 - ? C4H9 + CHO C4H9CHO
(3.5 g C4H10) (1 mol C4H10/58.12 g C4H10) (4 mol C/1 mol C4H10) (6.022 * 1023 C atoms/1 mol C) = 1.45 * 1023
Butane is C4H10. So; 100 grams C4H10 (1mol C4H10/58.12g)(4mol C/1mol C4H10 )(6.022 X 10^23/1mol C) = 4.14 X 10^24 atoms of Carbon.
C4H10 + O2 ------CO2 t H2O?
It is a molecular formula.This is because the formula C4H10 is not in its lowest ratio, which is C2H5
2C4H10 + 13O2 = 8CO2 + 10H2O
1.02g C4H10 /58.12 g/mol C4H10= .0175 mol C4H10 .0175 mol C4H10 * 10 mol H2O / 2 mol C4H10 = .0877 mol H2O .0877 mol H2O * 18.02 g H2O = 1.58 g H20 Final Answer: 1.58 g H2O
Butanes are organic molecules of the formula of C4H10. Butane has a molecular weight of 58.12 grams per mole.
The balanced equation is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O.
It is a hydrocarbon with the formula, C4H10
Butane is a gas.
C4H10 plus 9/2O2 equals 4CO plus 5H2O
Balance this combustion reaction first! 2C4H10 + 13O2 -> 8CO2 + 10H2O 0.86 moles C4H10 (13 moles O2/2 moles C4H10) = 5.6 moles of oxygen required ----------------------------------------
C4H10 has 2 isomers: butane and isobutane (also called 2-methyl propane)
14..4 elements of carbon,10 of hydrogen 2nd Answer: Good grief! There are 2 elements in C4H10: Carbon and Hydrogen.
The reaction is:2 C4H10 + 13 O2 → 8 CO2 + 10 H2OThe answer is 24 moles.
Butane Its also seen as Tetracarbon Decahydride
tetra-carbon deca-hydride
It is an empirical formula.
Yes, it is Dipole-dipole and london.
C4H10, the same as for butane.
No. Hydrocarbons are not miscible in water.
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