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3 years ago

8

Which element could provide one atom to make an ionic bond with fluorine? (Hydrogen will not make ionic bonds easily, so do not

use this element).

1 answer:

Ainat [17]3 years ago

6 0

Any element from group 1 of the Periodic Table.

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_______ are plants that produce seeds and flowers.

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angiosperms

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Aluminum can react with oxygen gas to produce aluminum oxide (Al2O3). What type of reaction is this? A. single replacement

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B. combustion would be the correct option

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Which carbon compound is soft, slippery, a decent electricity conductor, and bonded in a way that forms planes?

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Calculate the volume in liters of a barium acetate solution that contains of barium acetate . Be sure your answer has the correc

Ilya [14]

Answer:

1.09 L

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>

<em />

The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:

100 g × (1 mol/255.43 g) = 0.391 mol

0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:

0.391 mol × (1 L/0.360 mol) = 1.09 L

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3 years ago

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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:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

2 years ago