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3 years ago

Mg2OH has how many unique elements? A. 4 B. 2 C. 3 D. 5

Chemistry

1 answer:

frutty [35]3 years ago

5 0

Answer:

Magnesium hydroxide has 3 unique elements.

Explanation:

Long story short, 2 is the balancing compound in structure to make up hydroxide, therefore 1 compound would be left to create Mg(2O)H.

(This is only an opinion of mathematical science to me, I don't have complete understanding of this subject either, good luck.)

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PLEASE HELP!! Thanks! How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The

ZanzabumX [31]

Answer:

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$ , the melting point of water is $\rm 0\, ^{\circ} C$ , and that the boiling point of water is $\rm 100 \,^{\circ} C$ .

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$ .

Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$ . The change in temperature would be $\rm 10.0\,^{\circ}C$ .
Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
Step three: heat the 13.0 gram of water from $\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$ . The change in temperature would be $\rm 100\,^{\circ}C$ .
Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
Step five: heat the 13.0 gram of steam from $\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$ . The change in temperature would be $\rm 11.0\,^{\circ}C$ .

<h3>Energy required for step one, three, and five</h3>

The following equation gives the amount of energy $Q$ required to raise the temperature of an object by a $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

In this equation,

$c$ is the specific heat of this substance,
$m$ is the mass of the substance, and
$\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$ .

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$ .

<h3>Energy required for step two and four</h3>

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$ .

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$ .

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

H: 1.008,
O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$ .

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$ :

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$ .

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$

<h3>Energy required for all five steps, combined</h3>

$5.969\; \rm kJ + 33.7\; \rm kJ \approx 39.7\; \rm kJ$ .

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