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Andru [333]
3 years ago
15

The monatomic ions of Groups 1A(1) and 7A(17) are all singly charged. In what major way do they differ? Why?

Chemistry
2 answers:
grin007 [14]3 years ago
8 0

Answer:

Group 1A forms single charged cations whereas group 7A forms singly charged anions

Explanation:

Charge of an atom is indicative of the number of electrons lost or gained by the atom. Cations are positively charged ions formed by the loss of electrons while anions are negatively charged ions formed by the gain of electrons.

In the periodic table Group 1A belongs to the alkali metals which have a valence electron configuration of ns¹ i.e they have one electron in their valence orbital. Since they are metals these elements lose electrons readily gaining a +1 charge and forming a singly charged cation.

Group 7A belongs to the halogens which have a valence electron configuration of ns² np⁵ i.e they have 7 electrons in their valence orbital. They only need one more electron to complete their octet and become stable. As a result halogens are highly electronegative and form singly charged anion (-1 charge).

insens350 [35]3 years ago
3 0

<u>Answer:</u> Group 1 ions are known as cations and Group 17 ions are known as anions.

<u>Explanation:</u>

Ions are formed when an atom looses or gains electrons.

If an atom gains electrons, it leads to the formation of negative ions known as anions. <u>For Example:</u> Fluorine is a Group 17 element which gains 1 electron to form F^- ions.

If an atom looses electrons, it leads to the formation of positive ions known as cations. <u>For Example:</u> Sodium is a Group 1 element which looses 1 electron to form Na^+ ions.

Hence, group 1 ions are known as cations and Group 17 ions are known as anions.

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We are given:

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Calculating the moles of methanol in the solution, by using the equation:

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}

To calculate the mole fraction of methanol, we use the equation:

\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}

\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590

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\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}

\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

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