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Rus_ich [418]
2 years ago
9

the value of ksp for pbcl2 is 1.6. what is the lowest concentration of Cl- that would be needed to begin precipitation of PbCl2

in 0.010 M Pb(NO3)2
Chemistry
1 answer:
deff fn [24]2 years ago
7 0

Answer:

The minimum concentration of Cl⁻ that produces precipitation is 12.6M

Explanation:

The Ksp of PbCl₂ is expressed as:

PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)

The Ksp is:

Ksp = 1.6 = [Pb²⁺] [Cl⁻]²

When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.

A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:

1.6 = [0.010M] [Cl⁻]²

160 = [Cl⁻]²

12.6M =  [Cl⁻]

<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>
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How many moles of water as a gas can be formed 2.45 L
Alex17521 [72]

Answer:

0.11mole

Explanation:

Let us assume that the condition is at standard temperature and pressure(STP);

 Given parameters:

        Volume of water  = 2.45L

   Unknown:

       Number of moles found in this volume of water  = ?

Solution;

 At STP;

                 Number of moles  = \frac{volume of gas}{22.4}

 Input the parameters and solve;

                  Number of moles of water  = \frac{2.45}{22.4}   = 0.11mole

The number of moles of water found is  0.11mole

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How many pints are in 53 gallons'?
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Helpppp pleaseee ill give brainliest
Studentka2010 [4]

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ

5 0
2 years ago
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