Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
Answer:
0.80m of KOH
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, the solute is KOH and solvent is water.
Moles of 36g KOH -Molar mass: 56.1g/mol- are:
36g KOH × (1mol / 56.1g) = <em>0.642 moles of KOH</em>
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Now, as density of water is 1g/mL, mass of 800mL of water is:
800mL × (1g / mL) × (1kg / 1000g) = <em>0.800kg of water</em>
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Thus, molality is:
0.642moles of KOH / 0.800kg = <em>0.80m of KOH</em>
How many times has it halved?
24/2 = 12
12/2 = 6
6/2 = 3
It halved three times.
It halves once every 6 hours.
18 hours have passed.
Volume of the nitrogen gas = 49.8 L
<u>Explanation:</u>
It is given that the pressure, number of moles and temperature of nitrogen gas, and gas constant value being constant and it is taken as 0.08206 L atm mol⁻¹K⁻¹.
Temperature = T = 75°C = 75 + 273 = 348 K
Pressure = P = 0.992 atm
Number of moles = n = 1.73 moles
We have to use the ideal gas equation, PV = nRT, and rearranging the equation to get Volume in litres.
V = 
= 
= 49.8 L
So the volume of Nitrogen gas = 49.8 L