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Helen [10]
3 years ago
13

An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had

a carbon-14 decay rate of 14.8 disintegrations per minute per gram of carbon. Living organisms have a decay rate of 15.3 disintegrations per minute. How old is the bone?
Chemistry
1 answer:
Vlad [161]3 years ago
4 0

Answer : The time passed in years is 2.74\times 10^2\text{ years}

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}

t=274.64\text{ years}=2.74\times 10^2\text{ years}

Therefore, the time passed in years is 2.74\times 10^2\text{ years}

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How many grams of NaCl is present in .75 L of a 1 M solution of NaCl?
finlep [7]

Answer:

The answer to your question is 43.9 g of NaCl

Explanation:

Data

Grams of NaCl = ?

Volume = 0.75 L

Molarity = 1 M

Number of moles = ?

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

Number of moles = Molarity x volume

Substitution

Number of moles = 1 x 0.75

                             = 0.75

Molecular weight of NaCl = 23 + 35.5 = 58.5

Calculate the grams of NaCl using proportions

                           58.5 g of NaCl ---------------- 1 mol

                              x                     ---------------- 0.75 moles

                              x = (0.75 x 58.5) / 1

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6 0
3 years ago
Consider a galvanic cell composed of the SHE and a half-cell using the following reaction: Ag+(aq) + e− → Ag(s)
MissTica

Answer:

a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration

Explanation:

a. The two half cell reactions are

1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V

2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V

The balanced cell reaction is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V

b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

c. Use Nernst Equation

E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm

i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V

ii) E =  0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V

d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.

5 0
3 years ago
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