Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
Black coffee since it is a darker colour and has a heavier taste
Answer:
H+ + Cl- + OH- + Ca2+ -> Na+ + Cl- + H2O
Explanation:
<u>Answer:</u> The isomers are shown in the image below.
<u>Explanation:</u>
Isomers are defined as the chemical compounds having the same number and kinds of atoms but arrangement are different.
For the alkane having four carbon atoms and 1 bromine atom, the IUPAC name of the haloalkane is bromobutane
There are 4 possible isomers for the given haloalkane compound:
- 1-bromobutane
- 2-bromobutane
- 1-bromo-2-methylpropane
- 2-bromo-2-methylpropane
The isomers of the given organic compound is shown in the image below.
Answer:
Approximately 53.3 %.
Explanation:
Molar mass of C2H4O2
= 2 * 12.011 + 4 * 1.008 + 2 * 15.999
= 60.052
% oxygen = (31.998* 100) / 60.052 = 53.28 %
