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2 years ago

What is the main difference between sexual and asexual reproduction

Chemistry

2 answers:

andreyandreev [35.5K]2 years ago

5 0

Answer:

Asexual reproduction does not involve sex cells or fertilisation . Only one parent is required, unlike sexual reproduction which needs two parents. Since there is only one parent, there is no fusion of gametes and no mixing of genetic information.

Ahat [919]2 years ago

3 0

Answer:

Sexual - 2 Parents

Asexual - 1

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On January 28, 1986, the NASA shuttle orbiter mission STS-51-L and the tenth flight of Space Shuttle Challenger (OV-99) broke apart 73 seconds into its flight, killing all seven crew members, which consisted of five NASA astronauts and two payload specialists.

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If you change the 2 in front of 2O2 to a 3, what will be the change in the results on the right side of the equation? (1 point)

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There is an extra O2 molecule left over

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3 years ago

Which of the following are greenhouse gases? Select all that apply. Oxygen, carbon dioxide, nitrous oxide, nitrogen

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3 years ago

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Which of the following statements is true about gas particles?

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There is a lot of empty space between them

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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:

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

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3 years ago