Suppose you have a solution that might contain or all of the following cations: cu2+, ag+, ba2+, and mn2+. addition of hbr causes a precipitate to form. after the precipitate is filtered off, h2so4 is added to the supernate another precipitate forms. this precipitate is filtered off, and a solution of naoh is added to the supernatant liquid until it is strongly alkaline. no precipitate is formed. which ions are present in each of the precipitates? which cations are not present in the original solution?

Both Ba2+ and Ag+ are present.

Neither Cu2+ nor Mn^2+ are present.

Explanation:

Hello,

At first, the addition of HBr promotes the precipitation of AgBr as it is largely insoluble in water. Next, the addition of H2SO4 promotes the precipitation of BaSO4 as it is largely insoluble in water as well.

Finally, as no precipitate was formed due to the addition of sodium hydroxide and both Cu(OH)2 and Mn(OH)2, one concludes they aren't present into the solution.

Best regards.

6g h2o x (1 mole h2o / 18.0 g h2o) x (6.022x10^23 molecules / mole) = 1.20x10^24 molecules h2o 

11 g co2 x (1 mole co2 / 44.0 g co2) x (6.022x10^23 molecules / mole) = 1.51x10^23 molecules co2 

3.0 g co2 x (1 mole c / 12.0 g c) x (6.022x10^23 molecules / mole) = 1.5x10^23 atoms c 

caco3 has formula mass = 100 g/mole 
200 g caco3 x (1 mole / 100g) x (6.022x10^23 molecules / mole) = 1.20x10^24 formula units 

 
molar mass h2o = 2x1.0 + 1x16.0 = 18.0 g/mole 
molar mass f2 = 2x 19.0 = 38.0 g/mole 
molar mass cai2 = 40.1+ 2x126.9 = 294 g/mole 
molar mass pb(no3)2 = 207.2 + 2x(14.0 + 3x16.0) = 331 g/mole 

 
80.0 g h2o x (1 mole / 18.0 g) = 4.44 moles h2o 
45.0 g c6h12o6 x ( 1 mole / 180 g) = 0.25 moles c6h12o6 
22.0 g co2 x (1 mole / 44.0 g) = 0.500 moles co2 
56.0 g n2 x (1 mole / 28.0 g) = 2.00 moles n2 

 
2.0 moles lioh x (24.0 g/mole) = 48 g lioh 
5.0 moles ba(cn)2 x (189.3 g/mole) = 950 g ba(cn)2 
3.5 moles h2o x (18.0 g/mole) = 63 g h2o 
0.75 moles cuso4 x (159.6 g / mole) = 120 g cuso4 

 
2.0 moles he x (6.022x10^23 atoms / mole) = 1.2x10^24 he atoms 

3.0 moles nacl x (6.022x10^23 units nacl / mole) x (1 na+ / 1 unit) = 1.8x10^24 na+ ions 

0.25 moles ch4 x (6.022x10^23 molecules / mole) = 1.5x10^23 molecules ch4 

there are 3 atoms / 1 molecule of h2o 
1 mole x (6.022x10^23 molecules h2o / mole h2o) x (3 atoms / 1 mole h2o) = 1.8x10^24 atoms 

 
6.02x10^23 molecules h2o x (1 mole / 6.022x10^23 molecules) = 1.00 mole h2o 

3.01x10^23 molecules c2h6 x (1 mole / 6.022x10^23 molecules) = 0.500 moles c2h6 

1.2x10^24 molecules glucose x (1 mole / 6.022x10^23 molecules) = 2.0 moles glucose 

2.41x10^24 units cacl2 x (1 mole / 6.022x10^23 units) = 4.00 moles cac2 

 
1.20x10^24 molecules x (1 mole / 6.022x10^23 molecules) x (18.0 g / mole) = 36.0 g h2o 

2.41x10^24 units x (1 mole / 6.022x10^23 molecules) x (171.3 g / mole) = 342 g ba(oh)2 

1.51x10^22 atoms x (1 mole / 6.022x10^23 atoms) x (20.18 g/mole) = 0.503 g ne 

7.53x10^22 molecules x (1 mole / 6.022x10^23 molecules) x (16.04 g/mole) = 2.00 g ch4