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2 years ago

Complete the following radioactive decay problem. 210/84 Po --> 206/82 Pb + __/

Chemistry

1 answer:

vazorg [7]2 years ago

4 0

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

Your equation is:

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + ?

It becomes easier to balance the equation if we replace the "?" with an element symbol _x^yZ.

Then the equation becomes

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +_x^yZ

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Sum of superscripts: 210 = 206 + y, so y =4.

Sum of subscripts: 84 = 82 + x, so x = 2.

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

We should recall that $_2^4He$ is an α particle.

Thus, the equation represents the α decay of polonium-210 to lead-206.

The nuclear equation is

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

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A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .