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fredd [130]
2 years ago
9

Suppose that a 10-mL sample of a solution is to be tested for I− ion by addition of 1 drop (0.2 mL) of 0.13 M Pb(NO3)2.

Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
8 0

The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

<u>Given data : </u>

Volume of solution to be tested for I-ion = 10 mL

Volume of  Pb(NO₃)₂ = 0.2 mL

molarity of Pb(NO₃)₂ = 0.13 M

<h3>Determine the number of I that must be present </h3>

First step : calculate conc of PB²⁺ ions in the solution

conc of PB²⁺ ions =  ( molarity of Pb(NO₃)₂ * volume of  Pb(NO₃)₂ ) / ( total volume )

                              = ( 0.13 * 0.2 ) / ( 10 + 0.2 )

                              = ( 0.026 ) / ( 10.2 )  = 0.002549 M

Next step :<u> </u><u>determine the</u><u> molarity of  I</u>

using the dissociation reaction of PbI₂

PbI₂(s) ---> Pb²⁺ (aq)  + 2I (aq)

also;  Ksp = [ Pb²⁺ ] [ I ]²  ---- ( 1 )

From the question the given value of Ksp = 8.49 * 10⁻⁹

Therefore equation ( 1 ) becomes

8.49 * 10⁻⁹ = ( 0.002549 ) * [ I ]²

[ I ] = √ ( 8.49 * 10⁻⁹ ) / ( 0.002549 )

      = 0.0018 M

Final step :<u> Determine the minimum number of grams of I </u>

moles of I = molarity of I * total volume

                 = 0.0018 M * 10.2 mL

                 = 0.01836 * 10⁻³ mol

Hence we can conclude that The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Learn more about Pb(NO₃)₂ : brainly.com/question/25071409

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
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The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L =  1.600M

In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 -  0.595 = 0.005

[I2] =   1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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Type the correct answer in the box. Express your answer to three significant figures. Iron(II) chloride and sodium carbonate rea
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Answer:

The reaction can produce 287 grams of iron(II) carbonate

Explanation:

To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-

<em>Moles FeCl2:</em>

1.24L * (2.00mol / L) = 2.48 moles FeCl2

As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles

<em>Mass FeCO3:</em>

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<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>
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