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choli [55]
3 years ago
10

Metals react with ______ to form compounds that are alkaline.

Chemistry
2 answers:
kondor19780726 [428]3 years ago
7 0

Metals react with ______ to form compounds that are alkaline.

A. metalloids

B. oxygen (O)

C. non-metals

D. hydrogen (H)

The answer is D, Hydrogen (H).

kap26 [50]3 years ago
3 0

Answer: Option (B) is the correct answer.

Explanation:

When metals react oxygen atom then they tend to form metal oxides which are basic in nature. This is because when metal oxides are added in water then it results in the formation of metal hydroxides which are alkaline in nature.

For example, 2Mg + O_{2} \rightarrow 2MgO

Whereas when metals react with hydrogen atom then it results in the formation of metal hydrides which are acidic in nature.

When metals react with non-metals then it results in the formation of ionic compounds. These ionic compounds can be acidic or basic in nature.

Not all metals react with metalloids as only a few of them react with metalloids.

Thus, we can conclude that metals react with oxygen (O) to form compounds that are alkaline.

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Mass of N₂ will be formed : 50.12 g

<h3>Further explanation</h3>

Reaction

2 NH₃(g) + 3 CuO(s) → N₂(g) + 3 Cu(s) + 3 H₂O(g)

moles NH₃ = 3.58

ratio mol NH₃ : mol N₂ = 2 : 1

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Yakvenalex [24]

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a. There is little or no change in the entropy.

b. ΔS is greater than zero.

c. There is little or no change in the entropy.

d. ΔS is less than zero.

e. ΔS is greater than zero.

Explanation:

The change in the entropy (ΔS) depends on the change in the amount of gaseous moles, Δn(g) = ngas, products - ngas,reactants.

  • If Δn(g) > 0, the entropy increases (ΔS > 0)
  • If Δn(g) < 0, the entropy decreases (ΔS < 0)
  • If Δn(g) = 0, there is little or no change in the entropy.

<em>a. I₂(g) + Cl₂(g) → 2 ICl(g)</em>

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

<em>b. PCl₅(g) → PCl₃(g) + Cl₂(g)</em>

Δn(g) = 2 - 1 = 1. ΔS is greater than zero.

<em>c. CO₂(g) + H₂(g) → H₂O(g) + CO(g)</em>

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

<em>d. 2 CO(g) + 2 NO(g) → 2 CO₂(g) + N₂(g)</em>

Δn(g) = 3 - 4 = -1. ΔS is less than zero.

<em>e. 2 H₂O₂(l) → 2 H₂O(l) + O₂(g)</em>

Δn(g) = 1 - 0 = 1. ΔS is greater than zero.

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