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tekilochka [14]
3 years ago
13

Explain why a nuclear chain reaction is self-sustaining.

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
6 0

answer is written in word file attached below

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What two molecules were condensed in an aldol condensation to produce (ch3)3cch=chcoch3?
inessss [21]
The given compound is being synthesized by condensing Acetone and Pivaldehyde (Trimethylacetaldehyde).

First Acetone is treated with Base, the base abstracts the mildly acidic proton present at alpha position to carbonyl group. The resulting specie called enolate act as a nucleophile and attacks on highly reactive aldehyde which upon dehydration yields the Aldol Product. The Reaction is as follow,

5 0
4 years ago
One mole of which substance contains a total of 6.02 x 1023 atoms? (I need help please.)
Korolek [52]

Answer:

Oxygen

Explanation:

One mole of atoms of oxygen has a mass of 16 g, as 16 is the atomic weight of oxygen, and contains 6.02 X 1023 atoms of oxygen.

8 0
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\%.

3 0
3 years ago
You go to the same school as me BAHAHHAHA
Reptile [31]
Oh my i really hope no one from my school sees my comments
4 0
3 years ago
The structure of cyclopentane-1,2-diol​
Lady bird [3.3K]

Answer: five carbon atoms forms a ring. In adjacent two carbon atoms

An hydroxyl group (-OH) is attached. Molecular formula is

C5H8(OH)2

Explanation:

6 0
3 years ago
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