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Sindrei [870]
3 years ago
8

Can someone please help me

Chemistry
1 answer:
Vitek1552 [10]3 years ago
5 0
Of course i can help what do you need
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From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
The position of equilibrium would not be appreciably affected by changes in the volume of the container for
8090 [49]

Answer:

The position of equilibrium would not be appreciably affected by changes in the volume of the container for  NiO(s) + CO(g) ⇌ Ni(s) + CO2(g).

Correct Answer : Option A

Explanation:

The equilibrium position tends to change with increase or decrease in pressure or volume, or both of them. This happens because considering change in volume, when the volume of the container increases, the reactant molecule increases i.e. mole of gases and thus the position of equilibrium shifts towards the right side. Similarly in case of decreasing volume, reactant molecule decreases and the equilibrium position shifts left side.

And in case, when the mole of gases on both the sides of the equation i.e. reactant side and product side are equal, it will not have any effect on the equilibrium position on increasing or decreasing volume, or pressure.

4 0
3 years ago
What makes bullet holes wider on the exit side of a piece of glass?
denpristay [2]
The rounded shape of the bullet because as it goes through the glass it stops spinning and this make the hole wider
6 0
3 years ago
Read 2 more answers
Assume that you are given a solution of an unknown acid or base. How can you tell whether the unknown substance is acidic or bas
Jlenok [28]

Answer:

The litmus test.

Explanation:

Acids turn litmus paper red while bases turn it blue.

5 0
3 years ago
Read 2 more answers
Calculate the volumetric size of a water molecule in water vapor at normal conditions, assuming 1 mole of the vapor occupies 22.
Vedmedyk [2.9K]

Let volumetric size of a water molecule = v

Since, 1 mole of water consists of 6.023\times 10^{23} molecules of water.

Thus, total volume of 1 mole of water = volumetric size \times 6.023\times 10^{23}

Substitute the value of total volume of 1 mole of water i.e. 22.4 L in above formula.

22.4 L = v\times 6.023\times 10^{23}

v= \frac{22.4 L}{6.023\times 10^{23}}

Convert the unit litre to angstrom

v=\frac{22.4 L}{6.023\times 10^{23}}\times \frac{1dm^{3}}{1L}\times \frac{(10^{9})^{3}(A^{o})^{3}}{1 dm^{3}}

= 3.7\times 10^{-23}\times 10^{27} (A^{o})^{3}

= 3.7 \times 10^{4}(A^{o})^{3}

Therefore, the volumetric size of the water molecule is 3.7 \times 10^{4}(A^{o})^{3}


8 0
2 years ago
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