When iron rusts and forms iron oxide, the iron oxide has more mass than the iron. Which statement correctly explains this difference in mass?

The correct statement explaining the difference in mass when iron rusts to form iron oxide is:

Iron oxide has oxygen atoms, and pure iron does not.

When iron rusts, it reacts with oxygen in the presence of water or moisture in the environment, forming iron oxide (rust). The addition of oxygen atoms from the environment increases the total mass of the resulting compound compared to the original mass of the iron alone.

One molecule of calcium oxide, CaO, and one molecule of carbon dioxide, CO2, combine in a chemical reaction to form one substance. Given that matter is neither created nor destroyed, which chemical formula identifies the product of this reaction?

Ca2O3
CaCO2
CaCO3
Ca2Co3

In the reaction between one molecule of calcium oxide (CaO) and one molecule of carbon dioxide (CO₂), the product formed is calcium carbonate (CaCO₃).

Therefore, the correct chemical formula identifying the product of the reaction is:

CaCO3

Which statement correctly explains why the ashes of burnt paper have less mass than the original paper?

Other atoms combine with the paper to produce the ash.
Some of the atoms in the paper became gases.
Some of the atoms in the paper are destroyed when it burns.
The ash is created from new atoms that have less mass than the paper.

The correct statement that explains why the ashes of burnt paper have less mass than the original paper is:

Some of the atoms in the paper became gases.

When paper is burned, it undergoes a chemical reaction that converts some of its components into gases (such as carbon dioxide and water vapor), which are released into the atmosphere. This loss of gaseous products results in the ashes having less mass than the original paper.

It takes 1 molecule of nitrogen and 3 molecules of hydrogen to produce 2 molecules of ammonia using the following formula:
1N2 + 3H2 > 2NH3
Identify which combination of molecules will produce ammonia with no leftovers?

2N2 and 6H2
4N2 and 4H2
5N2 and 6H2
8N2 and 4H2

To determine which combination of molecules will produce ammonia (NH₃) with no leftovers, we can break down the reaction given:

1 N₂ + 3 H₂ → 2 NH₃

From this reaction, we see that:

• 1 molecule of nitrogen gas (N₂) reacts with 3 molecules of hydrogen gas (H₂) to produce 2 molecules of ammonia (NH₃).

Now, we need to identify which combination of molecules maintains the same ratio (1 N₂ : 3 H₂) while ensuring there are no leftovers.

Let's analyze each option:

1. 2N₂ and 6H₂:

   • Using 2 N₂ would require 2 × 3 = 6 H₂.
   • This matches perfectly, producing 2 × 2 = 4 NH₃. (No leftovers)

2. 4N₂ and 4H₂:

   • Using 4 N₂ would require 4 × 3 = 12 H₂.
   • 4 H₂ is insufficient to react with 4 N₂. (Leftovers)

3. 5N₂ and 6H₂:

   • Using 5 N₂ would require 5 × 3 = 15 H₂.
   • 6 H₂ is insufficient to react with 5 N₂. (Leftovers)

4. 8N₂ and 4H₂:

   • Using 8 N₂ would require 8 × 3 = 24 H₂.
   • 4 H₂ is insufficient to react with 8 N₂. (Leftovers)

The only option that uses the correct ratio of nitrogen to hydrogen with no leftovers is:

2N₂ and 6H₂.

in the equation
2H2 _ 102 >> 2H2O
if you change the 2H2 to 3H2 molecules and balance the equation, identify which statement is correct?

It is not possible to produce water without leftover molecules.
The result of the equation will be the same.
The equation will produce three molecules of water.
Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules.

To analyze the given reaction:

The original equation is:

\[ 2H_2 + O_2 \rightarrow 2H_2O \]

If we change \(2H_2\) to \(3H_2\), we need to balance the equation. The modified equation would look like this:

\[ 3H_2 + O_2 \rightarrow ? \]

To balance this equation correctly, we determine how many water molecules (H₂O) would be produced. Each water molecule contains 2 hydrogen atoms, so if we have 3 hydrogen molecules \( (3H_2)\), we have:

• 3 molecules of \(H_2\) provide \(3 \times 2 = 6\) hydrogen atoms.

To balance the equation with oxygen:

1. We need enough oxygen to react with 6 hydrogen atoms.
2. Since each water molecule contains one oxygen atom, to use 6 hydrogen atoms, we can produce:

\[ \frac{6 \text{ H}}{2} = 3 \text{ H}_2O \]

So the balanced equation would be:

\[ 3H_2 + 3O_2 \rightarrow 3H_2O \]

Now, let's evaluate the statements:

1. It is not possible to produce water without leftover molecules.

   • This statement is incorrect because the equation can be balanced without leftovers.

2. The result of the equation will be the same.

   • This statement is incorrect because changing to \(3H_2\) changes the amount of produced water.

3. The equation will produce three molecules of water.

   • This statement is correct. The balanced reaction with \(3H_2\) produces \(3H_2O\).

4. Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules.

   • This statement is misleading in this case because, with \(3H_2\), \(3 O_2\) are not used (only 1 O₂ molecule is needed to produce 3 water molecules).

Thus, the correct statement is:

The equation will produce three molecules of water.

"The equation will produce three molecules of water" isn't the right answer, it's "it is not possible to produce water without leftover molecules"

Let’s carefully analyze the situation regarding the chemical equation while making the adjustment you specified.

Starting with the original equation:

\[ 2H_2 + O_2 \rightarrow 2H_2O \]

When you change \(2H_2\) to \(3H_2\), you should balance the equation as follows:

\[ 3H_2 + O_2 \rightarrow ? \]

1. Hydrogen: 3 moles of \(H_2\) provide \(3 \times 2 = 6\) hydrogen atoms.
2. Water Production: To balance, we need to have enough oxygen to react with those 6 hydrogen atoms. Each water molecule (H₂O) contains one oxygen atom; hence, to produce 3 H₂O, we need:

\[ 3H_2 + 3O_2 \rightarrow 3H_2O \]

However, the calculation of oxygen shows that:

• To produce 3 molecules of water, you need 1.5 molecules of \(O_2\) (not 3), which cannot happen in whole numbers if you’re starting with whole molecules.

When looking at the balanced reaction:

• If you start with \(3H_2\), you can only use \(1.5 O_2\), meaning you would need to adjust your initial quantities of reactants. Since \(1.5 O_2\) isn't a whole number, there would indeed be an excess of hydrogen (leftover) if you only consider \(3H_2\) and \(1 O_2\).

Given this analysis, you are correct.

The accurate statement is:

"It is not possible to produce water without leftover molecules."

Thank you for your patience, and I apologize for the confusion in my previous response!