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3 years ago

What else is produced during the replacement reaction of magnesium and hydrochloric acid?

Chemistry

2 answers:

Sindrei [870]3 years ago

7 0

Answer: Magnesium Chloride $(MgCl_2)$ is produced other than hydrogen gas in the following reaction.

Explanation:

The given reaction is a type of single displacement reaction. This reaction is defined as the reaction in which a more reactive metal displaces a less reactive metal in a chemical reaction.

For the reaction of magnesium and hydrochloric acid, magnesium metal is more reactive than hydrogen, so it will easily displace hydrogen from its chemical reaction.

The equation follows:

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

By Stoichiometry,

1 mole of magnesium metal reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and hydrogen gas.

Hence, the other product formed in the given reaction is magnesium chloride.

erastova [34]3 years ago

4 0

Mg+ 2HCl ➡️ H2+ MgCl2

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Determine the total time that must elapse until only ¼ of an original sample of the radioisotope Rn-222 remains unchanged.

Arte-miy333 [17]

Answer:

7.6 days

Explanation:

Radon is a radioactive element and Radon-222 is it's most stable isotope. The half-life of Radon-222 has been found to be approximately 3.8 days.

Let, the initial amount of the Rn-222 = 1 = A

Final amount = $\frac{1}{4}$ = A'

We will use the following relation for calculating time elapsed in the decay

$A' = A(\frac{1}{2} )^\frac{t}{t_1_/_2} }$

Thus,

$\frac{1}{4} =1(\frac{1}{2} )^\frac{t}{3.8}$

We can write is as,

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2 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

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