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1 year ago

Nts

Chemistry

1 answer:

a_sh-v [17]1 year ago

6 0

The reaction is a synthesis reaction.

<h3>What are synthesis reactions?</h3>

Synthesis reactions are one of the numerous reactions in chemistry. These kinds of reactions involve the combination of two atoms of different elements resulting in the formation of new compounds.

Synthesis reactions are sometimes referred to as combination reactions, simply because they have to do with the combination of two or more atoms of different elements into a single compound.

The different elements that combine are referred to as the reactants while the result of the combination is known as the product.

For example: A + B --> AB

In the illustrated reaction, N2 combined with H2 to form NH3. N2 and H2 gases are the reactants while NH3 is the only product formed from the two reactants.

This is a good example of a synthesis/combination reaction.

More on synthesis reactions can be found here: brainly.com/question/24936069

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2 years ago

2000 minutes to hours dimensional analysis

elena-14-01-66 [18.8K]

Answer:

33.33 h

Explanation:

You know that

1 h = 60 min

If you divide both sides by 60 min, you get the conversion factor: 1 h/60 min = 1.

If you divide both sides by 1 h, you get the conversion factor: 1 = 60 min/1 h.

Both are conversion factors because they both equal one and multiplying a measurement by one does not change its value.

You choose the conversion factor that gives you the correct dimensions for your answer. It must have the correct dimensions on top (in the numerator),

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2 years ago

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.90 moles of magnesium perchlorate, Mg(ClO4)2.]\

Leviafan [203]

Answer:

4.90 moles of $Mg(ClO_4)_2$ will produce (9.8) moles of $Cl^{-}$ ,

(4.90) moles of $Mg^{2+}$ and

(39.2) moles of $O^{2-}$

Explanation:

From the question we are told that

The number of moles of is $n = 4.90 \ mols$

The formation reaction of $Mg(ClO_4)_2$ is

$Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2$

From the reaction we see that

1 mole of $Mg(ClO_4)_2$ is formed by 2 moles of $Cl^{-}$ 1 mole of $Mg^{2+}$ and 4 $O^{2-}$

This implies that

4.90 moles of $Mg(ClO_4)_2$ will produce (2 * 4.90) moles of $Cl^{-}$ ,

(1 * 4.90) moles of $Mg^{2+}$ and

(8 * 4.90) moles of $O^{2-}$

4.90 moles of $Mg(ClO_4)_2$ will produce (9.8) moles of $Cl^{-}$ ,

(4.90) moles of $Mg^{2+}$ and

(39.2) moles of $O^{2-}$

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