An atom has 24 protons, 22 electrons, and 28 neutrons. What is the ATOMIC # of the atom?
1 answer:
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Answer:
d. Oxidation and reduction
Explanation:
For this question we have to remember the definition of each type of reaction:
-) <u>Hydrogenation</u>
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In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.
-) <u>Alkylation</u>
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In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.
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-) <u>Hydrolysis</u>
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In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.
-) <u>Halogenation</u>
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In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.
-) <u>Ammoniation</u>
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In this reaction, we have the addition of the ammonium ion (). In the example, the ammonium ion is added to an acid.
-) <u>Oxidation and reduction</u>
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In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:
Reduction
Oxidation
Answer:
Al³⁺, S²⁻; Al₂S₃
Explanation:
Al is a metal, so it loses its three valence electrons in a reaction to form Al³⁺ ions.
S is a nonmetal. It needs two more electrons to complete its octet, so it tends to form S²⁻ ions.
We can use the criss-cross method to work out the formula of the compound.
The steps are
- Write the symbols of the anion and cation.
- Criss-cross the numbers of the charges to become the subscripts of the other ion.
- Write the formula with the new subscripts.
- Divide the subscripts by their highest common factor.
- Omit all subscripts that are 1.
When we use this method with Al³⁺ and S²⁻, the formula for the compound formed becomes Al₂S₃ .
