Answer:
[NH₄NO₃] at D → 0.279 M
Explanation:
This exercise involves a series of dilutions one after the other.
First of all, we calcualte ammonium nitrate's concentration at A.
15.71 g . 1 mol/ 80 g = 0.196 mol / 0.150 mL = 1.31 M
At B → 1.31 M . 20 mL/ 75 mL = 0.349 M
At C → 0.349 M . 15 mL / 25 mL = 0.209 M
[NH₄NO₃] at B = 0.349 M
[NH₄NO₃] at C = 0.209 M
So let's calculate the new moles
In 1 mL of B we have 0.349 mmoles
In 10 mL of B we have 3.49 mmoles
In 1 mL of C we have 0.209 mmoles
In 10 mL of C we have 2.09 mmoles
Volume of D = 10 ml + 10ml = 20 mL
Total mmmoles = 3.49 mmoles + 2.09 mmoles = 5.58 mmoles
[NH₄NO₃] at D = 5.58 mmoles / 20mL → 0.279 M
Answer : The number of moles of gas are in container 2 are 12.008 moles.
Explanation :
According to the Avogadro's Law, the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.
or,
where,
= initial volume of gas in container 1 = 1.30 L
= final volume of gas in container 1 = 2.33 L
= initial moles of gas in container 2 = 6.70 mole
= final moles of gas in container 2 = ?
Now put all the given values in the above formula, we get the final moles of the gas.
Therefore, the number of moles of gas are in container 2 are 12.008 moles.
D. Red and Violet are reflected, Green is absorbed
They are called isotopes. One example is Carbon-14, which has 2 extra neutrons.
Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).