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2 years ago

How many atoms are in 3.75 moles of Fe

Chemistry

1 answer:

denis23 [38]2 years ago

6 0

Answer:

2.26 × 10²⁴ atoms of Fe

Explanation:

To calculate the amount of atoms of a substance using the amount of moles, we must multiply by Avagadro's number.

(3.75mol) × (6.02 × 10²³) = 2.26 × 10²⁴ atoms of Fe

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2 years ago

Here is a list of ingredients for a simple cake:

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3 years ago

Explain the relationship between an individual and a Society

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2 years ago

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

$Molecular\ weight = \frac{Mass\ in\ g}{Moles}$

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2 years ago

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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2 years ago