How many atoms are in cuso4?

Which of the following is not a way populations are described?

gladu [14]

The last answer

the place where the organisms live

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4 0

1 year ago

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Which type of rock forms as a result of cooling and crystallization?

Dovator [93]

Answer:

Igneous rock is the type of rock that forms as a result of cooling an crystallization

Explanation:

Rocks are either sedimentary, igneous or metamorphic based on the chemical composition and the mode of formation.
Igneous rocks are formed as a result of the cooling and crystallization of magma from volcanic eruptions.
Igneous rocks are either intrusive, which form deep in the crust or extrusive which form on the surface of the earth.
Sedimentary rocks are formed as a result of deposition, and cementation of sediments.
Metamorphic rocks, on the other hand, are formed from the sedimentary and igneous rocks during the rock cycle.

7 0

3 years ago

Based on the following reactants:

Doss [256]

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5 0

3 years ago

Which model failed to explain the

11Alexandr11 [23.1K]

Answer:

C). The Bohr-Rutherford model

Explanation:

The 'Bohr-Rutherford model' of the atom failed to elaborate on the attraction between some substances. It essentially targeted hydrogen atoms and failed to explain its stability across multi-electrons. The nature and processes of the chemical reactions remained unillustrated and thus, this is the key drawback of this model. Thus, <u>option C</u> is the correct answer.

5 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

2 years ago