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telo118 [61]
2 years ago
12

The coin of silver (ag) having 8.5 grams weight.calculate the number of moles of silver in coin

Chemistry
1 answer:
AfilCa [17]2 years ago
6 0

0.0788 will be the number of moles of silver in coin.

<h3><u>How to find the number of moles?</u></h3>

A mole is the mass of a material made up of the same number of fundamental components. Atoms in a 12 gram example are identical to 12C. Depending on the material, the fundamental units may be molecules, atoms, or formula units.

A mole fraction shows how many chemical elements are present. The value of 6.023 x 10²³ is equivalent to one mole of any material (Avagadro's number). It can be used to quantify the chemical reaction's byproducts. The symbol for the unit is mol.

The number of moles formula is denoted by the following expression:

Number of moles = Mass of substance/mass of one mole

To view more about number of moles, refer to:

brainly.com/question/14080043

#SPJ4

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3 years ago
The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye
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Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

ln \frac{N}{N_0}  = - k*t ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

T _\frac{1}{2} = \frac{ln2}{k} ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

Multiply both side by k

28.8 yrs * k = \frac{ln 2 }{k} * k

Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

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