8. Which gas contributes the largest part of air? *

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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Because of the tilt of the Earth’s axis.

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How fast is an airplane traveling that accelerates at a rate of 20 mph per second for 30 seconds?

Nostrana [21]

Should be 600! hope this helps.

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Which of these makes it easier for scientists to understand shared measurement data? A. Having a universal language B.having a s

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0.010 moles of carbon C4H10 reacts with oxygon as in the queation 1.76g of carbon dioxide and 0.90 of water are produced. Use th

Sonbull [250]

The coefficients are 2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

Step 1. Gather all the information in one place.

M_r: 58.12 32.00 44.01 18.02

aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O

m/g: 1.76 0.90

n/mol: 0.010

Step 2. Calculate the mass of C₄H₁₀.

Mass = 0.010 mol C₄H₁₀ × (58.12 g C₄H₁₀/1 mol C₄H₁₀) = 0.581 g C₄H₁₀

Step 3. Calculate the mass of O₂

Mass of C₄H₁₀ + mass of O₂ = mass of CO₂ + mass of H₂O

0.581 g + x g = 1.76 g + 0.90 g

x = 1.76 + 0.90 - 0.581 = 2.079

Our information now has the form:

M_r: 58.12 32.00 44.01 18.02

aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O

m/g: 0.581 2.079 1.76 0.90

n/mol: 0.010

Step 4. Calculate the moles of each compound.

Moles of O₂ = 2.079 g O₂ × (1 mol O₂/32.00 g O₂) = 0.064 97 mol O₂

Moles of CO₂ = 1.76 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 0.040 00 mol CO₂

Moles of H₂O = 0.90 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.0499 mol H₂O

Our information now has the form:

aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O

n/mol: 0.010 0.064 97 0.040 00 0.0499

Step 5: Calculate the molar ratios of all the compounds.

a:b:c:d = 0.010:0.064 97:0.040 00:0.0499 = 1:6.497:4.000:4.99

= 2 :12.99:8.00:9.98 ≈ 2:13:8:10

∴ a = 2; b = 13; c = 8; d = 10

The balanced equation is

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

7 0

3 years ago