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Galina-37 [17]
3 years ago
15

What is a naturally occurring solid, inorganic compound or element?

Chemistry
1 answer:
Oksana_A [137]3 years ago
3 0
Answer is: <span>naturally occurring solid, inorganic compound or element is mineral.
Minerals usually have </span>crystalline form, that means than crystal constituents (atoms<span>, </span>molecules <span>or </span>ions<span>) are arranged in a highly ordered microscopic structure, forming a </span>crystal lattice<span> that extends in all directions.
</span><span>For example m</span>agnetite<span> is a </span><span>mineral of iron oxide.</span>
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Answer:

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Nylon is a polymer which is termed as
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Nylon is named as the most useful synthetic material. Nylon belongs to a family of materials called "polyamides."
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The type of bonding and the numbers of covalent bonds an atom can form with other atoms are determined by __________. View Avail
Dmitriy789 [7]

Answer:

The answer is: the number of unpaired valence electrons

Explanation:

The valence electrons are the number of electrons present in outer shell of an atom that participate in chemical bond formation, if the outer shell is not completely filled.

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State why vulcanologist make predications about when a volcano will erupt
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CaCO3(s) ∆→CaO(s) + CO2(g).
sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

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  • C: 12.011;
  • O: 15.999.

What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of CaCO₃ available:

\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}.

Molar mass of CO₂:

M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}.

Mass of the 0.400655 moles of \text{CO}_2 expected for the 40.1 grams of CaCO₃:

m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}.

What's the <em>percentage</em> yield of this reaction?

\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%.

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3 years ago
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