Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Answer:
Explanation:
Percent composition is percentage by the mass of element present in the compound.
The formula for chromium(III) nitrate is 
Molar mass of chromium(III) nitrate = 238.011 g/mol
1 mole of chromium(III) nitrate contains 9 moles of oxygen
Molar mass of oxygen = 16 g/mol
So, Mass= Molar mass*Moles = 16*9 g = 144 g
The answer:
we should know the meaning of each abbreviation:
ms means millisecond, its value is 10^-3 s
ns means means nanosecond, its value is 10^-9 s
ps means picosecond, its value is 10^-12 s
fs means femtosecond, its value is 1x 10^15 s
<span>Expressions of the quantity 556.2 x 10^-12 are</span>
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
556.2 x 10^-12 = 556.2 x 10^- 27 x 10^15 = 556.2 x 10^- 27 fs
Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
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