Use ideal gas law PV=nRT
Convert 5.00 atm to kPa since units must be relative to gas constant (r).
To do this multiply 5 by 101.03 (1 atm=101.3kPa)
Now plug in (506.5kPa)(10.0L)=n(8.31 L•atm/mol•K)(373K)
Solve for n (moles) to get approximately 1.634 mol. Now use dimensional analysis (1.634mol/1)(22.4L/1mol) = 36.6L
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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Answer:
change from a gas to aliquid
Explanation:
because gases expand when heated
The mass percentage composition of nitrogen in CO(NH2)2 is 46 .7%
calculation
% mass = molar mass of nitrogen/ molar mass of Co(NH2)2
molar mass of nitrogen = 14 x2 = 28g/mol since there is two nitrogen atom in Co(NH2)2\
molar mass of CO(NH2)2 = 12 + 16 + (14 x2) +( 2 x2)= 60 g/mol
the % mass is therefore =28 / 60 x100 =46.7%
