A student was holding a metal rod to a candle flame. After a while he noticed that his hand was getting hot.

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

Conduction

Explanation:

Conduction is when energy directly travels through a solid to get from one point to another.

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What is the concentration of h+ ions in a solution of hydrochloric acid that was prepared by diluting 35.0 ml of concentrated (1

Alekssandra [29.7K]

Hydrochloric acid ionisation is as follows;
HCl ---> H⁺ + Cl⁻
HCl is a strong base so there's complete dissociation of acid to H⁺ ions
The number of HCl moles is equivalent to number of H⁺ ions present
1 L of solution contains - 11.6 moles of H⁺ ions
In 35 ml number of moles - 11.6 mol/L / 1000 ml x 35 ml = 0.406 mol
This number of moles are dissolved in 500 ml
therefore molarity = 0.406 mol /500 ml x 1000 ml = 0.812 M

4 0

3 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:

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