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just olya [345]
3 years ago
11

Calculate the number of milliliters of 0.656 M KOH required to precipitate all of the Co2 ions in 187 mL of 0.745 M Co(NO3)2 sol

ution as Co(OH)2. The equation for the reaction is:
Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

THE MILLILITERS OF 0.656 M KOH REQUIRED TO PRECIPITATE ALL THE Co2 IONS IN 187 mL OF 0.745 M Co(NO3)2 SOLUTION IS 212.37 mL

EQUATION FOR THE REACTION IS :

2 KOH + Co(NO3)2 ----------> Co(OH)2 + 2 KNO3

Explanation:

Using dilution formula:

M1V1 = M2V2

V2 = M1 V1 / M2

M1 = 0.745 M

V1 = 187 mL

M2 = 0.656 M

V2 = unknown

V2 = 0.745 * 187 / 0.656

V2 = 139.315 / 0.656

V2 = 212.37 mL

the number of milliliters of 0.656 M KOH required to precipitate all of the Co 2 ions is 212.37 mL.

The equation for the reaction is:

2KOH + Co(NO3)2 ----------> Co(OH)2 + 2KNO3

That is 2 moles of potassium hydroxide react with 1 mole of cobalt(11) nitrate to form 1 mole of cobalt hydroxide and 2 moles of potassium nitrate

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2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
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<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

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