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3 years ago

How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14

Chemistry

1 answer:

Taya2010 [7]3 years ago

7 0

<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

Remaining mass of C-14 in a bone is 0.3125 g
Original mass of C-14 on the bone is 80.0 g
Half life of C-14 is 5370 years

We are required to determine the age of the bone;

Using the formula;

Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

= 8

Therefore, the number of half lives is 8
But, 1 half life is 5370 years
Therefore;

Age of the rock = 5370 years × 8

= 42960 years

Thus, the bone is 42960 years old

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Answer:

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Explanation:

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2 years ago

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

Starting with magnesium and oxygen atoms, use Lewis diagrams to show the formation of the

satela [25.4K]

Answer:

The electronic configuration of the magnesium atom shows that it has two unpaired outermost shell electrons while that of oxygen shows that it has six outermost shell electrons with two of them unpaired.

By means of elecron transfer, magnesium atom donates its two electrons to the oxygen atom to now have a charge of +2 while the oxygen then has a charge of -2. This results in an ionic bonding between the two atoms.

Explanation:

The structure is found in the attachment below