Answer:
MOLARITY = 0.9554 mol/dm3
Explanation:
525 grams of SnF2 in 3.50 L of wate
To obtain the molarity, calculate the concentration in g/dm3 and then concentration in mol/dm3
Concentration in g/dm3 = 525 g/ 3.50 L
= 525g/ 3.50 dm3
= 150g/dm3 of SnF4
Concnetration in mol/dm3
concentration in mol/dm3 = concentration in g/dm3 / Relative molecular mass of SnF2
RMM of SnF2 = ( 119 + 19*2) = 157 g/mol
= 150g/dm3 / 157g/mol
= 0.9554 mol/dm3 of solution.
The molarity of the solution is therefore 0.9554mol/dm3
Any element with an atomic number greater than 54.
Answer:
A... According to the periodic table, elements with one electron at the outermost shell or orbital are classified as alkali METALS
Convert 55.0g Ca(OH)2 to mols.
55.0g Ca(OH)2 = 0.742 mols Ca(OH)2
0.742mol Ca(OH)2/ 0.680M Ca(OH)2 = 1.09L Ca(OH)2
Neglecting the volume of the Ca(OH)2 itself, since it is minimal and its density wasn't provided, 1.09L would be the total volume of a 0.680M solution produced by dissolving 55.0g of Ca(OH)2 in enough water.