127.17 g LiCl x 1 mol/42.4 g x 6.02x10^23 Form.Units/moles = 1.81x10^24 Formula Units.
LiCl
Li2CrO4 * 8H2O
The correct formula for lithium chloride is LiCl. The subscript "2" is not needed because lithium only forms one positive charge and chloride only forms one negative charge.
You know because of solubility rules that LiCl disassociates 100% in water. So, knowing that molarity is equal to moles/liters Molarity LiCl = 1.97mol / 33.2 L Molarity = 0.059 M LiCl
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127.17 g LiCl x 1 mol/42.4 g x 6.02x10^23 Form.Units/moles = 1.81x10^24 Formula Units.
The formula mass of the ionic compound lithium chloride, LiCl is 6.9 + 35.5 = 42.4.Amount of LiCl = 98.2/42.4 = 2.32molThere are 2.32 moles of formula unit in a 98.2g pure sample of LiCl.To get the numerical number, multiply the quantity in moles by the Avogadro's constant.
42.394 grams.
To determine the number of molecules in a sample of LiCl, we need to first calculate the number of moles using the molar mass of LiCl (42.39 g/mol). Next, we use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. In this case, there are approximately (127.17 \text{ g} / 42.39 \text{ g/mol} \approx 3 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} ≈ 1.8 \times 10^{24}) molecules of LiCl in 127.17 g.
Formula: LiCl
LiCl
To find the number of molecules of LiCl in a 127.17 g sample, you first need to convert the mass of LiCl to moles using its molar mass. Then, use Avogadro's number (6.022 x 10^23) to convert moles to molecules. Calculate the number of molecules of LiCl in the sample using these values.
To calculate the number of molecules of LiCl in a 127.17g sample, you first need to determine the number of moles of LiCl in the sample using the molar mass of LiCl (6.94g/mol for Li and 35.45g/mol for Cl). Then, you can use Avogadro's number (6.022 x 10^23) to convert moles to molecules.
The salt lithium chloride is LiCl. It's an Li+ ion and a Cl- ion.
LiCl
LiCl
This is the chemical formula of sodium chloride.