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3 years ago

Predict the products of the following reaction: RbNO2 +BaCO3->

Chemistry

1 answer:

sukhopar [10]3 years ago

6 0

Answer:

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

Explanation:

The balanced reaction equation is shown below;

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

This reaction is possible because the reduction potential of Rb is -2.98V while that of Ba is –2.92 V. Hence Rb can displace Ba from its salt solution.

The equation is balanced since the number of atoms of each element on the left and right hand sides of the reaction equation are equal.

You might be interested in

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

$x = 0.0227 \ M$

3 0

3 years ago

CF4 + Br2 CBr4 + F2

luda_lava [24]

Answer:

Answers are in the explanation

Explanation:

Based on the reaction:

CF₄ + 2Br₂ → CBr₄ + 2F₂

The mole ratio of CF₄ is:

CF₄:Br₂ = 1:2

CF₄:CBr₄ = 1:1

CF₄:F₂ = 1:2

<em>Moles F2:</em>

Molar mass CF₄: 88.0g/mol

57.0g * (1mol / 88.0g) = 0.6477 moles CF₄ * (2mol F₂ / 1mol CBr₄) =

<h3>1.30 moles F₂</h3><h3 />

Molar mass CBr₄: 331.63g/mol

250.0g * (1mol / 331.63g) = 0.7539 moles CBr₄ * (2mol Br₂ / 1mol CF₄) =

1.51 moles Br₂ * (159.808g / mol) =

<em>Moles F2:</em>

4.8 moles CF₄ * (2mol F₂ / 1mol CF₄) =

<h3>9.6 moles F₂</h3><h3 />

<em />

5 0

3 years ago

What is the negatively charged sub-atomic particle of an atom?

vitfil [10]

C electron. Electrons have a negative charge!

7 0

3 years ago

Read 2 more answers

How many atoms are in 175.8 grams of Hg?

dybincka [34]

5.22*22^3 should be the answer

5 0

3 years ago

KClO3 (s) KCl (s) + 02 (g)<br><br><br>How do you balance this equation

Vitek1552 [10]

Answer:

2KCl(s) + 6O3O2

Explanation:

I think it's correct

7 0

3 years ago