How many moles is in 0.0688 Ag Cl
There's 4 moles.
1 mole Cl = 35.453g Cl 28.4g Cl x 1mol Cl/35.453g Cl = 0.801 mole Cl
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
There are 2 moles of Cl in 1 mole of CaCl2. The molar mass of Cl is 35.45 g/mol. So, in 435 g of CaCl2, there would be 2 moles of Cl, which is equal to 70.9 g of Cl.
To find the number of moles, divide the given mass of AgCl by its molar mass. The molar mass of AgCl is 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl). Therefore, 573.28 g ÷ 143.32 g/mol = 4 moles of AgCl.
0,5 moles Cl-
There's 4 moles.
it wouldn't be moles of Cl it would be Cl2 as chlorine doesn't exist as an atom it exsists as two joined to form a compound this is very easy stuff the answer is 15 moles of Cl2 as there 30 moles of Cl hope this helps learn your moles it's easy stuff
1 mole Cl = 35.453g Cl 28.4g Cl x 1mol Cl/35.453g Cl = 0.801 mole Cl
Mg2+ + 2 Cl- are in 1 : 2 ratio (of ions) so also 0.25 : 0.50 mole ratio
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
There are 2 moles of Cl in 1 mole of CaCl2. The molar mass of Cl is 35.45 g/mol. So, in 435 g of CaCl2, there would be 2 moles of Cl, which is equal to 70.9 g of Cl.
To find the number of moles, divide the given mass of AgCl by its molar mass. The molar mass of AgCl is 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl). Therefore, 573.28 g ÷ 143.32 g/mol = 4 moles of AgCl.
slightly
in the AgCl molecule two atoms present one is Ag (i.e.Ag in +1oxidation state) & another is Cl atom(Cl in -1).
To find the number of moles of PCl3, you need to first calculate the number of moles of Cl atoms in 3.68 * 10^25 atoms. There are 3 Cl atoms in each molecule of PCl3, so you divide the number of Cl atoms by 3 to get the number of moles of PCl3.
slightly