Which sentence about the periodic table of elements is true? *

1 answer:

5 0

Answer: metals are found on the left side of the periodic table

Justification:

A. All elements in period 2 are metals: FALSE.

The period 2 include metals and no metals: Li and Be are metals, while B is a metaloid, and C, N, O, and F are non-metals

B. All elements in group 18 are metals: FALSE.

The elements in group 18 are known as noble gases: He, Ne, Ar, Kr, Xe, Rn, and Og (recently named). All of them are non-metals.

C. Metals are found on the left side of the periodic table. TRUE.

With exception of H.

Alkalyne metals are the group 1 (Li, Na, K, Rb, Cs, and Fr).

Alkalyne-earth metals are the group 2: Be, Mg, Ca, Sr, Ba, and Ra.

Those two are the groups most to the left.

The non-metals are all on the right side of the periodic table.

D. Metals are found on the right side of the periodic table. FALSE.

According to the previous information.

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HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

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Answer: S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

On the reactant side:

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

On the product side:

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

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