Answer:
The molar mass (Mm) of the compound is 127.39 g/mole
Explanation:
ΔT = Kf. molality
ΔT = change in temperature = Tfinal - Tinitial = 2.8 - 5.5
Kf = freezing point depression constant = - 4.3 C/m (always negative because temperature is decreasing)
molality = moles of solute/Kg of solvent = mole (n)/(20 x 10^-3 Kg of benzene)
(2.8 - 5.5) = (-4.3) x molality
molality = 0.6279 mole/kg
0.6279 = mole of compound/(20 x 10^-3)
mole of compound = 0.01256 mole
mole (n) = mass (m) divided by Molar mass (Mm)
Molar mass = mass of compound / mole of compound
m/n = 1.6/0.01256 = 127.39 g/mole
Answer: C. Electrochemical cells involve oxidation-reduction reactions.
Explanation: Oxidation occurs at the anode, and reduction occurs at the cathode.
Explanation:
If the concentration of a substance is changed, the equilibrium will shift to minimise the effect of that change. If the concentration of a reactant is increased the equilibrium will shift in the direction of the reaction that uses the reactants, so that the reactant concentration decreases. For example, decreased volume and therefore increased concentration of both reactants and products for the following reaction at equilibrium will shift the system toward more products.
Answer:
The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.
Explanation:
Pressure in the container is P and volume is V.
Temperature of the helium gas molecules =
Molecules helium gas = x
Moles of helium has = 
PV = nRT (Ideal gas equation)
...[1]
After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.
Molecules of helium left after removal = 
Moles of helium has left after removal = 
...[2]
The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.
Answer:
C2H5NO
Explanation:
constituent elements N O C H
Mass composition 0.420 0.480 0.540 0.135
mole ratio 0.42/14 0.48/16 0.54/12 0.135/1
= 0.03 0.03 0.045 0.135
dividing by the smallest 0.03/0.03 0.03/0.03 0.045/0.03 0.135/0.03
ratio = 1 1 1.5 4.5
= 1 1 2 5
EMPERICAL FORMULA = C2H5NO
