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3 years ago

PLEASE HELP I HAVE TO DO THIS BENCHAMRK QUESTION!!!

Chemistry

1 answer:

GuDViN [60]3 years ago

8 0

PLEASE HELP I HAVE TO DO THIS BENCHAMRK QUESTION!!!

The chart shows parts of a plant and an animal, at different levels of organization. Consider the plant images. If we are referring to levels of organization in a plant which picture can BEST be used to fill in box Y?

first picture (A

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4 years ago

CaCO3(s) ∆→CaO(s) + CO2(g).

sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

Ca: 40.078;
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?

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

7 0

3 years ago