You would use this number, 6.02×1023 (Avogadro's number) to convert from particles, atoms, or molecules to moles. Whenever you go to the mole, divide by Avogadro's number. When you go to the unit from moles, multiply by Avogadro's number.
Answer:
energy is stored in the chemical bonds in molecules
Explanation:
Answer:
Kinetic energy increases.
Explanation:
In a solid, molecules do not have much room to move. They are very slow moving, which means that kinetic energy is low. In a liquid, molecules have more room to move. They are able to move faster than a solid, which means that kinetic energy is low.
From greatest to lowest kinetic energy:
gas, liquid, solid
430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.
<h3>What is Molarity?</h3>
- The amount of a substance in a specific volume of solution is known as its molarity (M).
- The number of moles of a solute per liter of a solution is known as molarity.
<h3>Calculation of Required amount of AgCl</h3>
Remember that mol/L is the unit of molarity (M).
We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.
0.75L×4.0M=3.0mol
Then, using the periodic table's molar mass for AgCl, convert from moles to grams:
3.0mol×143.321gmol=429.963g
The final step is to round to the correct significant figure, which in this case is two: 430g.
Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.
Learn more about Molarity here:
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Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁
