The formula mass of sodium chloride is 58.5
Amount of NaCl = 22.6/58.5 = 0.386mol
There are two moles of sodium ions in two moles of sodium chloride.
Every formula unit of sodium chloride has one sodium atom. Therefore, there are 4.0 moles of sodium ions in 4.0 moles of NaCl.
200/1000 x 6 = 1.2 moles
AgNO3 + NaCl ===> AgCl(s) + NaNO37 moles silver nitrate will produce 7 moles of silver chloride provided there is sufficient (at least 7 moles) of sodium chloride.
Molarity = moles of solute/Liters of solution ( find the moles ) 2.36 M NaCl = moles NaCl/5.08 Liters NaCl = 12.0 moles
0,40 moles of sodium chloride contain 23,376 g.
The answer is 0,175 moles.
There are 10 moles present in 585 g of sodium chloride.
23.3772 grams are there in four tenths moles of sodium chloride
3.6
.73 moles
There are two moles of sodium ions in two moles of sodium chloride.
20.67 grams NaCl (1mol NaCl/58.44 g NaCl) = 0.3537 moles of sodium chloride
This solution contain 0,3 mol of sodium chloride.
Every formula unit of sodium chloride has one sodium atom. Therefore, there are 4.0 moles of sodium ions in 4.0 moles of NaCl.
molecular formula for sodium chloride = NaClIf the mole (n) for NaCl = 5.3 moles, then the mole of sodium (Na) = 5.3 moles as well. 1 to 1 ratio mass = moles X molar mass m = 5.3 x 22.9 = 121.37 grams of sodium in 5.3 moles of sodium chloride
257g