Answer:
B) C3H3O and C6H6O2
Explanation:
Given data:
Molar mass of compound = 100 g/mol
Percentage of hydrogen = 5.45%
Percentage of carbon = 65.45%
Percentage of oxygen = 29.09%
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 5.45 / 1.01 = 5.4
Number of gram atoms of O = 29.09/ 16 = 1.8
Number of gram atoms of C = 65.45 / 12 = 5.5
Atomic ratio:
C : H : O
5.5/1.8 : 5.4/1.8 : 1.8/1.8
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03
n = 100 / 5503
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₃H₃O)
Molecular formula = C₆H₆O₂
So the whole point is that you need a abstract.
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
Answer:
a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.
b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.
c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.
d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling
temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.
