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MrRissso [65]
3 years ago
5

Which of the following descriptions of the storage requirements for alkali metals is the most accurate? Select the correct answe

r below: A. They must be kept separate from moisture and oxygen. B. They must be kept separate from other alkali metals. C. They must be kept separate from non-alkali metals. D. They must be kept separate from mineral oil.
Chemistry
1 answer:
elena55 [62]3 years ago
7 0

Answer:

A. They must be kept separate from moisture and oxygen

Explanation:

Alkali metals refers to a group of chemical elements in the periodic table. It is the common name given to Group 1 elements (excluding hydrogen). They are generally metallic in nature with physical properties like shining lustre, silvery in appearance etc. Alkali metals include lithium, Pottasium, Sodium, Rubidium etc.

One important chemical property of alkali metals is their high reactivity i.e. they tend to react quickly with other substances. For example, alkali metals react vigorously with water to form hydroxides. They also react with oxygen to form oxides. Due to this highly reactive properties, they are kept or stored away from moisture (water) and oxygen (present in air).

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Explanation:

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3 years ago
If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was
ANEK [815]

Answer:

Approximately 64\%.

Explanation:

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

The actual yield of \rm O_2 was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g).

Note the ratio between the coefficient of \rm H_2O\, (g) and \rm O_2\, (g):

\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}.

This ratio will be useful for finding the theoretical yield of \rm O_2\, (g).

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

  • \rm H: 1.008.
  • \rm O: 15.999.

Calculate the formula mass of \rm H_2O and \rm O_2:

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

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

Calculate the number of moles of molecules in 9.8\; \rm g of \rm H_2O:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}.

Make use of the ratio \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2} to find the theoretical yield of \rm O_2 (in terms of number of moles of molecules.)

\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})}  \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}.

Calculate the mass of that approximately 0.271996\; \rm mol of \rm O_2 (theoretical yield.)

\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}.

That would correspond to the theoretical yield of \rm O_2 (in term of the mass of the product.)

Given that the actual yield is 5.6\; \rm g, calculate the percentage yield:

\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}.

4 0
3 years ago
Rank the following salts in order of decreasing pH of their 0.1 M aqueous solutions.(a) FeCl2, FeCl3, MgCl2, KClO2 .(b) NH4Br, N
Firlakuza [10]

Answer:

a) FeCl2, FeCl3, MgCl2, KClO2.

KClO2 --> K+ + ClO2-; ClO2- will hydrolyse to form HClO +OH-

Mg+2, Fe+2 and Fe+3 ions will form acidic solutions, since theyfom slightly amount of

Mg+2 + 2H2O <-> Mg(OH)2 + 2H+

Fe+2 + 2H2O <-> Fe(OH)2 + 2H+

Fe+3 + 3H2O <-> Fe(OH)3 + 3H+

Therefore;

decreasing pH is high pH to low pH:

KClO2 > MgCl2 > FeCl2 > FeCl3

b) NH4Br, NaBrO2, NaBr, NaClO2.

NH4Br is acidic, forms NH4+ and NH4+ dnates H+ to form NH3 andH+

NaBrO2 is basic, forms Na+ + BrO2- then H2O + BrO2- HBrO2

NABr is neutral, NaClO2 is basics, forms Na+ + ClO2-then H2O + ClO2- HClO2

decreasig pH:

NaClO2 > NaBrO2 > NaBr >NH4Br

Note that HClO2 is stronger acid than HBrO2, therefore, expectmore HBrO2 formation

NaBrO2 > NaClO2 > NaBr >NH4Br

5 0
3 years ago
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