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Firdavs [7]
2 years ago
10

Which of the following, when added to an equilibrium mixture represented by the equlibrium below, will not alter the composition

of the original equilibrium mixture? Mg(OH)2(s) Mg2+(aq) + 2OH-(aq)
A. Addition of Mg(OH)2(s) to the equilibrium mixture.
B. Addition of HCl(aq) to the equilibrium mixture.
C. Addition of NaOH(s) to the equilibrium mixture.
D. Addition of Mg(NO3)2(s) to the equilibrium mixture.
E. Addition of Fe(NO3)3(aq) from the equilibrium mixture.
Chemistry
1 answer:
Kamila [148]2 years ago
8 0

Answer:

A. Addition of Mg(OH)2(s) to the equilibrium mixture.

Explanation:

A. Addition of Mg(OH)2(s) to the equilibrium mixture.  

The equilibrium will not change because if more solid is added, it will occur in the same original equilibrium and also, there is no change in the equilibrium if the solid and liquid states of the species are changed because there concentration hardly changes as they are present in large amounts.

B. Addition of HCl(aq) to the equilibrium mixture.  

HCl will furnish protons which will combine with the hydroxide ions and shifting the equilibrium in the forward direction.

C. Addition of NaOH(s) to the equilibrium mixture.  

It will increase the hydroxide ion concentration and thus shifting equilibrium backward.

D. Addition of Mg(NO3)2(s) to the equilibrium mixture.  

This will dissolve in the solution and furnish magnesium ions and thus shifting equilibrium backward.

E. Addition of Fe(NO3)3(aq) from the equilibrium mixture.

The ions it will form will form salts with their counter ions and thus changing the concentration of the products in the equilibrium.

<u>Hence, A is the answer.</u>

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We can predict the order of the elements given above according from the highest to lowest first ionization energies by using the trends in a periodic table. For elements in a family, the ionization energy decreases as it goes down. Therefore, the correct order would be Be, Mg, Ca, Sr.
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A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav
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Answer: 3

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Equation for the reaction of decay of _{19}^{40}\textrm{K} radioisotope follows:

Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles  

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_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e

By the stoichiometry of above reaction,

1 mole of _{40}^{91}\textrm{Zr} is produced by 1 mole _{41}^{91}\textrm{Nb}

So, 0.17 moles of _{40}^{91}\textrm{Zr} will be produced by = \frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}

Amount of _{82}^{212}\textrm{K}

decomposed will be = 0.17 moles

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a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = 0.024

a_o = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

0.024=\frac{0.194}{2^n}

n=3

Therefore, 3 half lives have passed.

3 0
2 years ago
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in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally
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Answer:

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Explanation:

Given data:

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Now we will put the values in formula.

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