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densk [106]
2 years ago
11

Zinc + Hydrochloric acid — Zinc chloride + Hydrogen [Zn +2 HCI → ZnCl2 + H2] is an

Chemistry
1 answer:
Jet001 [13]2 years ago
6 0
Double replacement i believe
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For the reaction COCl2(g)⇌CO(g)+Cl2(g), K= 2.19×10−10 at 373 K
vladimir1956 [14]

Answer:

D) The equilibrium lies far to the left

Explanation:

According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:

K = \frac{[CO][Cl_{2}]}{[COCl_{2}]} = 2.19×10^{-10}

([X] means = concentration of X)

This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.

6 0
3 years ago
What is the energy of an electron in the hydrogen atom determined by?
mezya [45]

Answer:

The energy of a hydrogen atom's electron is determined by which principal quantum number n value corresponds to the energy state the electron occupies. where n=1,2,3,... is the quantum number that quantizes the energy levels. That is, they are discrete energy values proportional to 1n2 .

Explanation:

6 0
3 years ago
The _______ are found on the right side of the arrow in a chemical reaction. A. reactants B. products C. subscripts D. coefficie
julia-pushkina [17]

Answer:

B. products are found on the write side of the arrow in a chemical reaction.

8 0
3 years ago
Read 2 more answers
How strong are bonds between subatomic particles?
Aleks [24]

Answer:

they are indeed very strong

8 0
3 years ago
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
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