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Shkiper50 [21]
3 years ago
6

A 0.180 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib

rium? Assume the formation constant of Ni ( NH 3 ) 2 + 6 is 5.5 × 10 8 .
Chemistry
1 answer:
Nutka1998 [239]3 years ago
4 0

Answer:

1.09 x 10⁻⁴ M

Explanation:

The equation of the reaction in given by

Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆

At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product

As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase

From the equation,

1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)

therefore

0.18 M of Ni would react with 1.08M  (6 x 0.18M) aqueous  NH₃ to give 0.18M of Ni(NH₃)6

At equilibrium,

1.08M of NH3 would have reacted to form the product  leaving

(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.

Therefore, formation constant  K which is the ratio of the concentration of the product to that of the reactant is given by

                       K  = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]

   5.5 x 10⁸          =  0.18 M / [Ni²⁺] [0.12]⁶

                   [Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)

                            =0.18 M / 0.00001642

                           = 1.09 x 10⁻⁴ M

[Ni]²⁺                  =   1.09 x 10⁻⁴ M

Hence the concentration of Ni²⁺ is  1.09 x 10⁻⁴ M

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A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
seropon [69]

<u>Answer:</u> The p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, M_1=(2\times 0.0031)=0.0062M

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, M_2=(2\times 0.0042)=0.0084M

To calculate the concentration of nitrate ions in the solution, we use the equation:

M=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Putting values in above equation, we get:

M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M

Calculating the p-function of zinc ions and nitrate ions in the solution:

  • <u>For zinc ions:</u>

\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]

\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51

  • <u>For nitrate ions:</u>

\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]

\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14

Hence, the p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

5 0
3 years ago
1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he
julsineya [31]

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, \Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

5 0
3 years ago
At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre
padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0
3 years ago
How many atoms are represented in 28 grand of sodium
Nady [450]
The molar mass of sodium is 22.99 ㅤ ㅤ ㅤ 22.99 (28) = 643.72 mol now multiply by avogrados number to find the number of atoms. ㅤ ㅤ ㅤ 643.72 (6.022 x 10^23) = 3.88 x 10^26 number of atoms
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3 years ago
Calculate the pH of the following?
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The correct answer is c
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