First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is
2 g/mol + 3 g/mol = 5 g/mol AB
for the case of A2B3
A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol
therefore A2B3 = 13 g/mol
<h2>Answer:</h2>
Moles of a gas = 0.500
Volume = 2.50 L
Pressure = 13. atm
Temperature = ?
Solution:
Formula:
PV = n RT
Putting the values in formula:
T = PV/nR = 13 * 2.5 / 0.5 * 0.082057
= 32.5/0.041 = 792.68 K
T = 792.68 K
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