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4 years ago

In the reaction 3H2 + N2 → ____ NH3, what coefficient should be placed in front of NH3 to balance the reaction?

1 answer:

5 0

2. 3H2 + N2 means you have 6 Hs and 2 Ns. NH3 has one N and 3 Hs, so you need 2 NH3s in order to have the 2 and 6 of each that you need on both sides of the reaction.

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Which of the following will increase the number of collisions without changing the energy of reactant molecules?

natulia [17]

Answer:increasing the concentration of reactants

Explanation:

Collision is the phenomenon in which the reactant molecules come to nearest closness,as a result the reactants are converted into products.

Now the number of effective collision is directly proportional to the number of reactants added..

6 0

3 years ago

Use the Bohr Model below to answer the questions:

Bas_tet [7]

Answer:

Ca(Calcium)

20 electrons

2 valence electrons

Explanation:

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2 years ago

What type of mirror should be used for this application? Explain why this mirror is the right choice for a telescope.

Artyom0805 [142]

Answer:

use a concaved miror

Explanation:

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3 years ago

Is sublimation exothermic or endothermic

Masja [62]

Answer:

endormic

Explanation:

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5 0

3 years ago

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

OleMash [197]

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

3 0

3 years ago