The answer is 62.00 g/mol.
Solution:
Knowing that the freezing point of water is 0°C, temperature change Δt is
Δt = 0C - (-1.23°C) = 1.23°C
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation
Δt = i Kf m
1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg)
x = 0.069435 mol
Therefore, the molar mass of the solute is
molar mass = 4.305g / 0.069435mol = 62.00 g/mol
The purpose of an universal indicator is to test wether a solution is acid or if its a base. It changes colors according to the PH's. It helps a lot in the indication of a chemical reaction because it can say if <span>each component loses or gains protons depending upon the acidity or basicity of the solution being tested.An universal indicator can say if a determined solution proves to be endothermic or exothermic. If the solution is not tested as acid or base then we cannot knwo if there will be an endothermic reaction or an exhotermic one</span>
Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.
Solution :
Using ideal gas equation,
where,
n = number of moles of gas
w = mass of ammonia gas = ?
P = pressure of the ammonia gas = 2.55 atm
T = temperature of the ammonia gas = 
M = molar mass of ammonia gas = 17 g/mole
R = gas constant = 0.0821 L.atm/mole.K
V = volume of ammonia gas = 3.00 L
Now put all the given values in the above equation, we get the mass of ammonia gas.
Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.
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
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