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Olenka [21]
3 years ago
10

Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c

hloride precipitate. What was the molar concentration of chloride ions in the solution (in mol/L)?
Chemistry
1 answer:
KiRa [710]3 years ago
3 0

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

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3 0
1 year ago
Read 2 more answers
1.
Leya [2.2K]

Answer:

V₂ = 530.5 mL

Explanation:

Given data:

Initial temperature = 20.0°C

Final temperature = 40.0 °C

Final volume = 585 mL

Initial volume = ?

Solution:

Initial temperature = 20.0°C (20+273 = 293 K)

Final temperature = 40.0 °C (40+273 = 323 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂  

V₂ = 585 mL × 293 K / 323 K

V₂ = 171405 mL.K / 323 K

V₂ = 530.5 mL

6 0
3 years ago
15 POINTS!!!! Please answer 1-8
Effectus [21]
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4 0
3 years ago
1) How much water must be added to 500.mL of 0.200 M HCI to produce a 0.150M solution?
zloy xaker [14]

Answer:

1)

(500.mL)(.200M)/.150M = 667 mL

667mL - 500mL = 167mL of water is needed

2)

1.0 L = 1000mL

M1 V1 = M2 V2

(1.6 mol/L) (175 mL) = (x)(1000mL)

x = .28M

6 0
3 years ago
How many moles of HCl are present in 4.7 L of a 4.23 M HCl solution?
REY [17]

Hey There!

Volume = 4.7 L

Molarity = 4.23 M

Therefore:

n = M *V

n = 4.23 * 4.7

n = 19.881 moles of HCl

hope this helps!

6 0
3 years ago
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