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4 years ago

What happens to an atom of sulfur (S) if it loses an electron?

Chemistry

2 answers:

alexira [117]4 years ago

4 0

Answer: Sulfur atom will form a cation on loosing electrons.

Explanation:

There are 3 subatomic particles present in an atom. They are: protons, electrons and neutrons.

Protons carry positive charge, electrons carry negative charge and neutrons does not carry any charge.

Any neutral atom has equal number of protons and electrons.

An ion is formed when atom looses or gain electron.

If an atom gains electrons, it will have more number of electrons than protons. Thus, it will carry negative charge and form an anion.

If an atom looses electrons, it will have less number of electrons than protons. Thus, it will carry positive charge and forms a cation.

It is given that sulfur atom looses an electron. So, this atom will have more number of protons than electrons. Thus, it will form a cation.

Hence, sulfur atom will form a cation on loosing electrons.

WITCHER [35]4 years ago

3 0

Answer:

An atom becomes anion when it gains or loses electrons. The ions that are formed when an atom loses electrons are positively charged because they have more protons in the nucleus than electrons in the electron cloud.

Explanation:

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How many atoms are below? C6H1206

Vinvika [58]

Answer:

24 atoms

Explanation:

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4 years ago

Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2

Mariana [72]

Answer: The value of $\Delta S^o$ for the reaction is 1051.93 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K$

Hence, the value of $\Delta S^o$ for the reaction is 1051.93 J/K

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3 years ago

Number of moles in 3.70 x 10^-1 g of boron

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The answer is 10.81 g of boron.

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