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wlad13 [49]
3 years ago
8

Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.

Chemistry
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)

These are the colligative properties:

ΔP = P° . Xm . i  →  Lowering vapor pressure

ΔT = Kb . m . i  → Boiling point elevation

ΔT = Kf . m . i  →  Freezing point depression

π = M . R . T  →  Osmotic pressure

Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.

CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:

CaCl₂  → Ca²⁺ + 2Cl⁻

We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3

KNO₃ → K⁺ + NO₃⁻

We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2

Option a, is the best.

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Lead (ll) nitride+ammonium sulfate > lead(ll) sulfate + ammonium nitride
Alenkinab [10]

Answer:

Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N

Explanation:

Let's rewrite the given word equation in its chemical balanced equation representation:

1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions: Pb_3N_2.

2. Ammonium sulfate consists of an ammonium cation with a 1+ charge and sulfate anion with a 2- charge, two ammonium cations needed: (NH_4)_2SO_4.

3. Lead(II) sulfate would have one lead cation and one sulfate anion, as they have the same magnitude of charges with opposite signs: PbSO_4.

4. Ammonium nitride would require three amonium cations to balance the nitride anion: (NH_4)_3N.

Let's write the balanced equation:

Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N

7 0
3 years ago
What do you think is happening at the nanoscale that keeps the salad dressing with an emulsifier mixed?
saveliy_v [14]

Answer:

✨ science ✨ and ✨ big brain stuff ✨

Explanation:

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4 0
3 years ago
Balloon has a volume of 600-ml at temperature of 360 K. If the temperature of
Molodets [167]

Answer:

V₂ ≈416.7 mL

Explanation:

This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.

  • V₁ / T₁ = V₂ / T₂

where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.

The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,

  • V₁= 600 mL
  • T₁= 360 K
  • T₂= 250 K

Substitute the values into the formula.

  • 600 mL /360 K = V₂ / 250 K

Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.

  • 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
  • 250 K * 600 mL/360 K = V₂

The units of Kelvin cancel, so we are left with the units of mL.

  • 250 * 600 mL/360=V₂
  • 416.666666667 mL= V₂

Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.

  • 416.7 mL ≈V₂

The volume of the balloon at 250 K is approximately 416.7 milliliters.

5 0
3 years ago
Prediction for Scandium (II) and Cl
SCORPION-xisa [38]

Answer:

(a) Eka-aluminum and gallium are two names of the same element as Eka-Aluminium has almost exactly the same properties as the actual properties of the gallium element. The properties: atomic mass, density, melting point, formula of chloride and formula of oxide are almost the same.

Explanation:

Scandium - Eka boron.

      (ii) Gallium - Eka aluminium.

      (iii) Germanium - Eka silicon.

3 0
3 years ago
Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation:
Leya [2.2K]
<span>1) Use the balanced chemical equation to find the molar ratios (proportions) of each product and reactant.

3N2H4(l)→4NH3(g)+N2(g)

=> molar ratios: 3 mol N2H4 : 4 mol NH3

2) Use the product to reactant molar ratio, and the quantity of reactant to determine the yield:

2.0 mol N2H4 * [4mol NH3] / [3mol N2H4] = 2*4/3 mol NH3 = 2.7 mol NH3

Answer: 2.7 mol
 
</span>
8 0
3 years ago
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