V=4.8 L
c=5.0 mol/L
M(Mg)=24.3 g/mol
1) n(HCl)=cv
2) m(Mg)=M(Mg)n(HCl)/2
3) m(Mg)=M(Mg)cv/2
m(Mg)=24.3*5*4.8/2=291.6 g
Answer: The density of silver metal will be ![10.50g/ml[\tex]Explanation:Density is defined as the mass contained per unit volume.[tex]Density=\frac{mass}{Volume}](https://tex.z-dn.net/?f=10.50g%2Fml%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EDensity%20is%20defined%20as%20the%20mass%20contained%20per%20unit%20volume.%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7Bmass%7D%7BVolume%7D)
Given : Mass of silver = 194.3 grams
Volume of silver= volume of water displaced= ![260.5-242.0=18.5mltex]Putting in the values we get:[tex]Density=\frac{194.3g}{18.5ml}=10.50g/ml/tex]Thus density of silver metal will be [tex]10.50g/ml](https://tex.z-dn.net/?f=260.5-242.0%3D18.5mltex%5D%3C%2Fp%3E%3Cp%3EPutting%20in%20the%20values%20we%20get%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7B194.3g%7D%7B18.5ml%7D%3D10.50g%2Fml%2Ftex%5D%3C%2Fp%3E%3Cp%3EThus%20density%20of%20silver%20metal%20will%20be%20%5Btex%5D10.50g%2Fml)
When the solutions are diluted in half:
So [ Ca+2] = 0.001/2 = 0.0005 M
[SO4-2]= 0.01/2 =0.005
when Q = [Ca2+][SO4-2]
so by substitution:
Q = 0.0005*0.005 = 2.5x10^-6
by comparing with Ksp value
So when Q < Ksp
∴ the answer is no precipitation will occur, as the solution is unsaturated.