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2 years ago

Which process produces the energy that is used by solar panels?

Chemistry

2 answers:

Setler [38]2 years ago

8 0

Answer:

Explanation: got it right

KengaRu [80]2 years ago

6 0

Answer:

Explanation:

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3 years ago

Read 2 more answers

a solution of KCl in water has a concentration of 0.243 M. The solution has a volume of 0.580 L. How many grams of KCl are prese

Novosadov [1.4K]

Answer:

10.5 g

Explanation:

Step 1: Given data

Molar concentration of the solution (C): 0.243 M

Volume of solution (V): 0.580 L

Step 2: Calculate the moles of solute (n)

Molarity is equal to the moles of solute divided by the liters of solution.

M = n/V

n = M × V

n = 0.243 mol/L × 0.580 L = 0.141 mol

Step 3: Calculate the mass corresponding to 0.141 moles of KCl

The molar mass of KCl is 74.55 g/mol.

0.141 mol × 74.55 g/mol = 10.5 g

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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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What would a chemist most likely study about a car? 1- The amount of rotations of the wheel that occur in 1 mile. 2- The reactio

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