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3 years ago

Give the nuclear symbol for the isotope of silicon for which a=29? Enter the nuclear symbol for the isotope (e.G., 42he).

1 answer:

7 0

The mass number is the summation of number of proton and neutron present in a nucleus of an atom. For the neutral atom the number of positive charge (number of proton) must be equal to the number of electrons. The number of electrons present in an atom is the atomic number of the atom. The standard way to express the mass number (a) and atomic number (m) of a atom (say X) is $x^{a}_{m}$ . Now for silicon number of electron or atomic number is 14. And the mass number (a) given 29. Thus the expression nucleus of silicon will be $Si^{29}_{14}$

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Why the noble gases tend not to react

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How many total ions are in 2.95 g of magnesium sulfite?

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Answer:

Approximately $5.65 \times 10^{-2}\; \rm mol$ , which is approximately $3.41 \times 10^{22}$ particles.

Explanation:

Formula of magnesium sulfite: $\rm MgSO_3$ .

Look up the relative atomic mass of $\rm Mg$ . $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm Mg$ : $24.305$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

The ionic compound $\rm MgSO_3$ consist of magnesium ions $\rm {Mg}^{2+}$ and sulfite ions $\rm {SO_3}^{2-}$ .

Notice that $\rm {Mg}^{2+}\!$ and $\rm {SO_3}^{2-}\!$ combine at a one-to-one ratio to form the neural compound $\rm MgSO_3$ . Therefore, each $\rm MgSO_3\!$ formula unit would include one $\rm {Mg}^{2+}$ ion and one $\rm {SO_3}^{2-}$ ion (that would be two ions in total).

Calculate the formula mass of one such formula unit:

$\begin{aligned}&\; M(\mathrm{MgSO_3}) \\ = & \; 24.305 + 32.06 + 3 \times 15.999 \\ = & \; 104.362\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the mass of one mole of $\rm MgSO_3$ formula units (which includes one mole of $\rm {Mg}^{2+}$ ions and one mole of $\rm {SO_3}^{2-}$ ions) would be $104.362\; \rm g$ .

Calculate the number of moles of such formula units in that $2.95\; \rm g$ of $\rm MgSO_3$ :

$\begin{aligned}n&= \frac{m}{M}\\ &=\frac{2.95\; \rm g}{104.362\; \rm g \cdot mol^{-1}} \approx 2.82670 \times 10^{-2}\; \rm mol\end{aligned}$ .

There are two moles of ions in each mole of $\rm MgSO_3$ formula units. Therefore, that $2.82670 \times 10^{-2}\; \rm mol$ of $\rm MgSO_3\!$ formula units would include approximately $2 \times 2.82670 \times 10^{-2}\; \rm mol \approx 5.65\times 10^{-2}\; \rm mol$ of ions.