Answer:
Explanation:
Molarity is a measure of concentration in moles per liter.
The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.
- molarity= 1.2 mol NaNO₃/L
- liters of solution=0.25 L
- moles of solute =x
We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.
The units of liters cancel, so we are left with the units moles of sodium nitrate.
There are 0.3 moles of sodium nitrate.
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
Answer:
-6.4x10⁻¹⁹ C
Explanation:
The elementary charge of one electron is -1.60x10⁻¹⁹C, so each electron has its charge, and a sample with more than one electrons will have a multiple of its charge, which is proportional to the number of electrons. So, if the oil droplet had 4 electrons, thus the charge will be four times the elementary charge:
4*(-1.60x10⁻¹⁹) C = -6.4x10⁻¹⁹ C
Answer:
Explanation:
1. To identify what's been oxidised and what's been reduced in a redox reaction
2. In naming compounds
3. To work out reacting proportions in titration reactions
