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3 years ago

Hi All,

Chemistry

1 answer:

Novosadov [1.4K]3 years ago

4 0

Answer:Combination.

Decomposition.

Single displacement.

Double displacement.

Combustion.

Redox.

Explanation:

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One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0

3 years ago

20 moles of NH3 are needed to produce ? Moles of H2O

AfilCa [17]

Hi :)

20 mol NH3 x 6 H2O/4 NH3 = 30 mol H2O

Hope this helped :)

5 0

3 years ago

PLEASE HELP!?!.!,! What are the products of the combustion of a hydrocarbon?

Tatiana [17]

Answer:

carbon dioxide and water

Explanation:

Example: Combustion of Methane (CH₄(g))

CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**

____________________

Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,

Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)

Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)

Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)

The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*

______________________

*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.

Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)

=> 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g) <= Standard Form of Rxn

Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)

Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)

=> 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn

______________________

**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).

3 0

3 years ago

Read 2 more answers

How many digits to the right of the decimal point should be used to report the result?

aalyn [17]

Answer:

In engineering and science the common stand is two places.

For example if you get a calculation of 4.567 round up and give the result of 4.57

4 0

3 years ago

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

Kaylis [27]

Molecular equation:
$H_{2}SO_{4}(aq) + 2NaOH(aq)--\ \textgreater \ Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Complete ionic equation:

2H^{+}+SO_{4}^{2-} + 2Na^{+} +2OH^{-}--\ \textgreater \ 2Na^{+}+SO_{4}^{2-} + 2H_{2}O(l)


Near 2H^{+},SO_{4}^{2-} , 2Na^{+} , OH^{-} 

should be written (aq).

Remove ions that do not change from both sides

2H^{+} + 2OH^{-}=2H_{2}O

Net ionic equation:

H^{+}(aq) + OH^{-}(aq)=H_{2}O(l)

Answer is D.

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8 0

3 years ago