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3 years ago

What is the volume of liquid in this graduated cylinder? Be sure your answer has a reasonable number of decimal places.

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Answer:

76 ml

Explanation:

The complete question is attached in the image.

The volume of a liquid is the amount of space that the liquid occupies or contains. Since for a liquid, it takes the shape of the container, when measuring the volume, we just measure the space occupied by the liquid in the container.

From the image, when determining the volume, we are to use the value of the lower meniscus. Hence the volume of the liquid is gotten to be 76 ml

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What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

Marta_Voda [28]

Answer:

The value is $x = 0.0227 \ M$

Explanation:

From the question we are told that

The concentration of $KCN \ \ i.e \ \ CN^{-}$ is $M_1 = 0.091 \ M$

The solubility product constant for $NiS$ is $K_{sp} = 3.0 *10^{-19}$

The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

Generally the dissociation reaction for NiS is

$Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

$4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

$4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=> $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=> $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=> $119.8 - 5264x =\sqrt{x}$

Square both sides

$(119.8 - 5264x)^2 =x$

=> $14352.04 - 1261255 x + 27709696x^2 = 0$

=> $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

The value of x is $x = 0.0227 \ M$

Hence the amount in terms of molarity (concentration) of $Ni(CN)_4 ^{2-}$ and $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

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3 years ago

A West Virginia coal is burned at a rate of 8.02 kg/s. The coal has a sulfur content of 4.40 % and the bottom ash contains 2.80

aleksley [76]

Answer: The annual emission rate of SO2 is 1.08 × $10^{7}$ kg/yr

Explanation:

The rate <em>r</em> at which the coal is been burnt is 8.02 kg/s.

Amount of sulphur in the burning coal is given as 4.40 %

i.e., 4.4/100 × 8.02 = 0.353 kg/s. Which is equivalent to the rate at which the sulphur is been burnt.

Since the burning of sulphur oxidizes it to produce SO2, it follows that the non-oxidized portion of the sulphur will go with the bottom ash.

The bottom ash is said to contain 2.80 % of the input sulphur.

Hence the portion of the SO2 produced is 100 — 2.80 = 97.20 %.

The rate of the SO2 produced is percentage of SO2 × rate at high sulphur is been burnt.

= 97.20/100 × 0.353 kg/s.

= 0.343 kg/s.

To get the annual emission rate of SO2, we convert the kg/s into kg/yr.

1 kg/s = 1 kg/s × (60 × 60 × 24 × 365) s/yr

1 kg/s = 31536000 kg/yr

Therefore, 0.343 kg/s = 0.343 × 31536000 kg/yr

= 10816848 kg/yr

= 1.08 × 10^7 kg/yryr.

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