Answer:
17.1195 grams of nitric acid are produced.
Explanation:

Moles of nitrogen dioxide :

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.
Then 0.5434 moles of nitrogen dioxides will give:
of nitric acid.
Mass of 0.3623 moles of nitric acid :

Theoretical yield = 22.8260 g
Experimental yield = ?


Experimental yield of nitric acid = 17.1195 g
The mass stays the same because if you have the same amount of steam then it can't change. The volume will get slightly smaller because the average kinetic energy of the molecules is less, so they move around less, so they take up less space. The particles are moving less fast.
Answer:
A synthesis reaction is <em>a reaction that occurs when two or more reactants combine into one product.</em>
Explanation:
A reaction that occurs when two or more reactants combine into one product is called a synthesis reaction.
A reaction that occurs when one element within a compound is exchanged with another element is called a single replacement reaction.
A reaction that occurs when a substance combines with molecular oxygen, releasing light and energy is called combustion.
A reaction that occurs when a single substance breaks apart and forms two or more new substances is called decomposition.
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
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