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2 years ago

Which atom would tend to gain 1 valence electrons from another atom in order to become stable

Chemistry

1 answer:

Mrrafil [7]2 years ago

4 0

Answer: Option (D) is the correct answer.

Explanation:

Atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7.

So, in order to attain stability it tends to gain one electron from a donor atom. Therefore, a chlorine ion is formed as $Cl^{-1}$ .

As it gains only one electron that is why, -1 charge occurs.

Whereas Ar is a noble gas with atomic number 18. As it has completely filled octet so it will be unreactive in nature as it is already stable.

Similarly, He is also a noble gas so it is unreactive in nature.

Atomic number of boron (B) is 5 with electronic distribution 2, 3. So, in order to attain stability it needs to lose 3 electrons.

Thus, we can conclude that Cl is the atom which tends to gain 1 valence electron from another atom in order to become stable.

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M = 1 + 2(12) + 3(1) + 2(16) = 60 g

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Glucose, C6H12O6

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3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

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