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2 years ago

PLS HELP ASAP I WILL GOVE BRAINLYEST

Chemistry

1 answer:

Radda [10]2 years ago

5 0

Answer:

4 genetically different daughter cells

Explanation:

in image

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Fofino [41]

Hah what is this work

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3 years ago

When a mixture containing cations of Analytical Groups I–III is treated with H2S in acidic solution, which cations are expected

Fantom [35]

Answer:

Analytical Groups I and II

Explanation:

Precipitation reactions happen when anions and cations in aqueous solution mix together to form an insoluble ionic solid which is now referred to as a precipitate.

Whether or not a type of reaction like that takes place can be determined by utilizing the solubility principles for common ionic solids.

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2 years ago

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How many electrons does it take to fill the outer energy level of most atoms

Orlov [11]

Depends on the element it can by up to 3, 8, or maybe 16.

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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:

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

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2 years ago

A semiconductor:

Alex

Answer:

D. has characteristics between conductor and insulator

7 0

1 year ago