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attashe74 [19]
3 years ago
12

A main group metal was studied and found to exhibit the following properties: It does not occur free in nature. It loses valence

electrons readily. It reacts readily with the halogens, oxygen, and nitrogen. It is produced by high-temperature reduction of its oxide. It reacts vigorously with water. Which of the following metals might have been the element studied?
1-Li2-Sr3-Ba4-Al5-Na
Chemistry
1 answer:
ANEK [815]3 years ago
3 0

Answer:

Barium (Ba)

Explanation:

Barium is a group 2 elements.

Valence shell electronic configuration = [Xe]6s^2

Its valence shell has only two electrons and by loosing these electrons, it attains inert gas configuration. therefore, it looses electron easily.

It is highly reactive and because of its high reactive nature, never present in nature in free state.

In general, reactivity increases down the group in the periodic table.

This is because down the group atomic size increases and electrons becomes far from the nucleus, so outer electrons are not tightly held by the nucleus. Therefore, valence shell electrons loose easily which in turn increases reactivity.

So, Ba is more reactive among all the given elements.

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For the following reaction, 5.61 grams of carbon monoxide are mixed with excess water . Assume that the percent yield of carbon
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<u>Answer:</u> The ideal yield of carbon dioxide is 7.506 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of carbon monoxide = 5.61 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol

The chemical equation for the reaction of carbon monoxide and water follows:

CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)

By Stoichiometry of the reaction:

1 mole of carbon monoxide produces 1 mole of carbon dioxide

So, 0.200 moles of carbon monoxide will produce = \frac{1}{1}\times 0.200=0.200mol of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.200 moles

Putting values in equation 1, we get:

0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g

To calculate the experimental yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Percentage yield of carbon dioxide = 85.3 %

Theoretical yield of carbon dioxide = 8.8 g

Putting values in above equation, we get:

85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g

Hence, the ideal yield of carbon dioxide is 7.506 grams

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Answer:

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