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-Dominant- [34]
3 years ago
7

Which of the following is an example of a synthesis reaction?

Chemistry
2 answers:
Setler79 [48]3 years ago
8 0
A, because a synthesis reaction is A+B—> AB

B is a decomposition AB—> A+B (opposite of synthesis)
C is acids and base acid 1 + base 1 —> acid 2 + base 2
D is double displacement AB+CD —> AD+CB
Neko [114]3 years ago
8 0
A because I know so trust me on this one Willis
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4 years ago
A 0.130 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
rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of Ni^{2+} ions at equilibrium is, 4.31\times 10^{-8}

Explanation :  Given,

Moles of NiCl_2 = 0.130 mol

Volume of solution = 1 L

Concentration=\frac{Moles }{Volume}

Concentration of NiCl_2 = Concentration of Ni^{2+} = 0.130 M

Concentration of NH_3 = 1.20 M

K_f=5.5\times 10^8

The equilibrium reaction will be:

                      Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}

Initial conc.    0.130       1.20                     0

At eqm.             x       [1.20-6(0.130)]      0.130

                                   = 0.42

The expression for equilibrium constant is:

K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}

Now put all the given values in this expression, we get:

5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}

x=4.31\times 10^{-8}

Thus, the concentration of Ni^{2+} ions at equilibrium is, 4.31\times 10^{-8}

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