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nadezda [96]
3 years ago
15

BRAINLIEST IF ANSWERED CORRECTLY WITH EXPLANATION! HELP PLEASE:)

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

First question

Pb(NO_3)_2 + KI \rightarrow PbI_2 + KNO_3

Balaneced:

Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3

Find how many moles potassium iodide it is in 345.0 grams

n_{KI} = \frac{345.0 \ g}{166.0 \ g/mol} = 2.078 \ mole

Find how many moles potassium nitrate there is for 2.078 moles KI

n_{KNO_3} = \frac{2}{2} \cdot n_{KI} = 2.078 \ mole

Find how many grams potassium nitrate there is in 2.078 moles KNO_3

m_{KNO_3} = 2.078 \ mol \cdot 101.1 \ g/mol = 210.1 \ g

Second question

yield \ \% = \frac{experimental}{theoretical} \cdot 100 = \frac{195.3 \ grams}{210.1 \ grams} \cdot 100 = 93.0 \ \%

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Could someone explain how they got this answer, explain step by step plz
gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

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allsm [11]

The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.

Concentration is defined as the number of moles of a solute present in the specific volume of a solution.

According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.

M₁V₁=M₂V₂

Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml

Rearrange the formula for M₂

M₂=(M₁V₁/V₂)

Plug all the values in the formula

M₂=(1.0M×14 ml/25 ml)

M₂=14 M/25

M₂=0.56 M

Therefore, the concentration of a dextrose solution after the dilution is 0.56M.

To know more about dilution

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