Answer:
In this conditions, the gaswll weight 46.74 g.
Explanation:
The idal gas law states that:
PV = nRT,
P: pressure = 740 mmHg = 0.97 atm
V: volume = 14.5 L
n: number of moles
R: gas constant =0.08205 L.atm/mol.K
T: temperature = 29°C = 302.15K

1 mol gas ___ 82 g
0.57 mol gas __ x
x = 46.74 g
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
Answer:
13.5 %
Explanation:
First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:
- 500 mL * 1.00 g/mL = 500 g
Then we <u>calculate the mass percent of potassium sulfate</u>, using the formula:
Mass of Potassium Sulfate / Total Mass * 100%
- 78 g / (78 + 500) g * 100 % = 13.5 %