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4 years ago

Why does h2so4 break up to h+ and hso4-?

1 answer:

3 0

Sulfuric acid or H2SO4 is a strong acid and when in aqueous solution it dissociates completely into hydronium ions or H+ and to HSO4- ions. The sulfuric acid can donate H+ decreasing the pH of a solution. Hope this answers the question.

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Some Reactions are endothermic

Alinara [238K]

Answer:

endothermic means energy is taken in and temperature increases.

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4 years ago

If you have 5 unknowns and they were secondary alcohol aldehyde, carboxylic acid, amine, and alkane Put plan to identify the unk

Finger [1]

Answer:

the unknowns

Explanation:

7 0

3 years ago

¿Quién estableció el concepto moderno de elemento?

Zina [86]

Answer:

Robert Boyle

Explanation:

En muchos textos se suele considerar a Robert Boyle como el científico que introdujo en la Química un concepto de elemento diferente al empleado por los aristotélicos o por los alquimistas y que sirvió de antecedente al que formulara Lavoisier en 1789.

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3 years ago

Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.

Pavlova-9 [17]

Answer: $5.81\times 10^6J/mol$

Explanation:

Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.

$C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles$

Thus $1.577\times 10^{-4}moles$ of sucrose releases = 916.6 J of heat

1 mole of sucrose releases = $\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J$ of heat

Thus ∆H value for the combustion reaction is $5.81\times 10^6J/mol$

6 0

4 years ago

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

4 years ago