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3 years ago

How do you name branched-chain alkanes?

Chemistry

1 answer:

tankabanditka [31]3 years ago

7 0

Answer:

Explanation:

Branches or side chains are called alkyl radicals, they are named substituting the termination -ano for -yl or -il when it is part of a hydrocarbon.

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What occurs when the nuclei of two very small atoms combine into one large nucleus?

Svetlanka [38]

Answer:

The combination of nuclei to form a bigger and heavier nucleus is known as Fussion. The consequence of fusion is the absorption or release of energy.

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4 years ago

You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar

melisa1 [442]

since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.

3 0

3 years ago

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{175\ g}{101.543\ g/mol}$

$Moles_{copper(II)\ fluoride}= 1.7234\ mol$

Also,

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,,

$Volume =\frac{Moles\ of\ solute}{Molarity}$

Given, Molarity = 0.00231 M

So,

$Volume =\frac{1.7234}{0.00231}\ L$

<u>Volume = 746 L</u>

7 0

3 years ago

How many grams of iron contain 4.06X10^24 atoms of iron?

Oxana [17]

4.06x20^24/6.02x10^23 = 6.744 moles x 55.845 g/mole = 376.61868grams

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:

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

3 0

3 years ago