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For the equation

[tex]\[ 3 \text{Cu} + 8 \text{HNO}_3 \rightarrow 3 \text{Cu}(\text{NO}_3)_2 + 2 \text{NO} + 4 \text{H}_2\text{O}, \][/tex]

how many units of \(\text{NO}_3\) are represented on the products side?

A. 2
B. 3
C. 6
D. 8

Sagot :

GinnyAnswer
Let's analyze the chemical equation given:

[tex]\[ 3 Cu + 8 HNO_3 \rightarrow 3 Cu(NO_3)_2 + 2 NO + 4 H_2O \][/tex]

We need to determine the number of nitrate ions (\(\text{NO}_3\)) on the products side of the reaction.

First, we focus on the product \( \text{Cu(NO}_3\text{)}_2 \):
- Each molecule of \( \text{Cu(NO}_3\text{)}_2 \) contains 2 nitrate ions (\(\text{NO}_3\)).

Given in the equation, there are 3 molecules of \( \text{Cu(NO}_3\text{)}_2 \):
- So, each of these 3 molecules has 2 \(\text{NO}_3\) ions.

To find the total number of nitrate ions on the products side, multiply the number of \(\text{NO}_3\) ions in one molecule by the number of these molecules:

[tex]\[ 3 \text{ molecules} \times 2 \text{ NO}_3 \text{ per molecule} = 6 \text{ NO}_3 \][/tex]

Thus, the total number of [tex]\(\text{NO}_3\)[/tex] ions on the products side is [tex]\(\boxed{6}\)[/tex].
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