Fazendo distribuição eletronica de titânio e possui número atomico que possui número atomico 22 qual o subnivel mais energético

Nconcentration refers to the molar concentration of an ion in solution. it may be identical to, or greater or less than, the molar concentration of the compound containing the ion that was used to make the solution. for soluble salts, the molarity of a particular ion is equal to the molarity of that compound times the subscript for that ion. for example, 1 m of alcl3 is 1 m in al3+ and 3 m in cl−. 1 m of (nh4)2so4 is 2 m in nh4+ and 1 m in so42−. part a what is the concentration of k+ in 0.15 m of k2s? view available hint(s) nothing m m part b if cacl2 is dissolved in water, what can be said about the concentration of the ca2+ ion? view available hint(s) if is dissolved in water, what can be said about the concentration of the ion? it has the same concentration as the cl− ion. its concentration is half that of the cl− ion. its concentration is twice that of the cl− ion. its concentration is one-third that of the cl− ion. part c a scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. how many grams of na3po4 will be needed to produce 550 ml of a solution that has a concentration of na+ ions of 0.700 m ? express your answer numerically in grams. view available hint(s) mass of na3po4 n a 3 p o 4 = nothing g provide feedback

A5.000 g sample of niso4 h2o decomposed to give 2.755 g of anhydrous niso4. what is the formula of the hydrate? what is the full chemical name for the hydrate? what is the molar mass of the hydrate? niso4•_h2o what is the mass % of water in the hydrate?

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes