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VLD [36.1K]
2 years ago
9

10.0 mL of a 0.100 mol L–1 solution of a metal ion M2+ is mixed with 10.0 mL of a 0.100 mol L–1 solution of a substance L. The f

ollowing equilibrium is established:
M2+(aq) + 2L(aq) Picture ML22+(aq)

At equilibrium the concentration of L is found to be 0.0100 mol L–1. What is the equilibrium concentration of ML22+, in mol L–1?

Someone Help Me
Chemistry
2 answers:
LenKa [72]2 years ago
8 0

<u>Answer:</u> The concentration of ML_2^{2+} at equilibrium is 0.045 M.

<u>Explanation:</u>

We are given:

[M^{2+}]_{initial}=0.100M

[L]_{initial}=0.100M

For the given chemical equation:

                        M^{2+}(aq.)+2L(aq.)\rightleftharpoons ML_2^{2+}(aq.)

Initially:             0.100M        0.100M

At eqllm:          0.100 - x      0.100 - 2x          x

We are also given:

[L]_{eqllm}=0.0100M

Equating the two values:

0.0100=0.100-2x\\\\x=0.045M

Hence, the concentration of ML_2^{2+} at equilibrium is 0.045 M.

Mrac [35]2 years ago
7 0
The chemical reaction is
<span>                                 M2+(aq) + 2L(aq) <==> ML22+(aq)
</span>Intial concentration   0.10           0.10        
Change                       -x              -2x               +x
Equilibrium                0.10 - x    0.01 = 0.10 - 2x  x
Solving for x
0.01 = 0.10 - 2x
x = 0.045
The equilibrium concentration of ML22+ is 0.045 mol L-1


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Genrish500 [490]

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q = m * ΔHf

the mass of ice m = 65 g

the heat of fusion of water at 0C = ΔHf = 334 J/g

Therefore: q = 65 g * 334 J/g = 21710 J

Now:

4.184 J = 1 cal

which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal

Hence the heat required is 5188.8 cal or 5.2 Kcal (approx)

5 0
2 years ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
rusak2 [61]

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

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3 0
2 years ago
For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o
galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

  • MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

          S                   S              2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

  • NaF(s) → Na+(aq)  +  F-(aq)

        0.10M     0.10M        0.10M

  • MgF2(s) ↔ Mg2+(aq)  + 2F-(aq)

          S'                 S'              2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility  is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=PCl_5%20%5Crightleftarrows%20PCl_3%20%2B%20Cl_2" id="TexFormula1" title="PCl_5 \rightleftarrow
Butoxors [25]

<u>Answer:</u> The total pressure of the container will be 2.00 atm

<u>Explanation:</u>

We are given:

Initial moles of phosphorus pentachloride = 1.00 atm

For the given chemical reaction:

PCl_5\rightleftharpoons PCl_3+Cl_2

By Stoichiometry of the reaction:

1 mole of PCl_5 produces 1 mole of PCl_3 and 1 mole of chlorine gas

So, 1.00 atm of PCl_5 will also produce 1.00 atm of PCl_3 and 1.00 atm of chlorine gas when the reaction goes to completion.

Total pressure of the container when the reaction goes to completion  = 1.00 + 1.00 = 2.00 atm

Hence, the total pressure of the container will be 2.00 atm

3 0
2 years ago
A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0
2 years ago
Read 2 more answers
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