Since the compound has 1.38 time that of oxygen gas at the same conditions of temperature and pressure, we have the relationship:
MW/MWoxygen = 1.38
MW = 44.16
Since there is water formed during the reaction, the formula of the compound must be:
XaHb
where a and b are the coefficients of each element.
If the compound reactions with oxygen forming water and an oxide of the element X, the combustion reaction must be:
XaHb + ((2a + (b/2))/2) O2 = a (XO2) + (b/2)(H2O)
Using dimensional analysis:
10 (1/44.16) (b/2 / 1) (18) = 16.3
Solving for b:
b = 8
The compound now is XaH8. Most probably, the compound is C3H8 since it has a molecular formula of 44 and it reacts with O2 to form water and CO2.
Answer: 0.83mol/m3
Explanation:
Number of mole = 2.5 mmol = 0.0025mol
Recall
1L = 0.001m3
Therefore 3L = 3x0.001 = 0.003m3
0.003m3 contains 0.0025mol
Therefore, 1m3 will contain = 0.0025/0.003 = 0.83mol
The concentration in mol/m3 is 0.83mol/m3
Answer:
2-methylebutane
Explanation:
it is a hydrocarbon and there is a specific Rule for its nomenclature.
that is known as IUPAC naming system. According to it
- The compound that has all single bonds name ends on ane.
- Names according to the number of Carbons,
* For one carbon compound it is known as methane,
* For 2 ethane,
* For 3 propane
* For 4 Butane and so on.
- select the longest chain of compound
- Start numbering with lowest substituted position at carbon
- branch compound name as Iso and neo as whole
So keeping in mind the compound can be named as
2-methylebutane
- As the longest chain contain four carbon atoms so it will name as butane
- The substitution of methyl group is on C-2
*Note: it can also named as isopantane as it is branched and have five carbons but it could not be 3-isopentane.