The balanced complete ionic equation for precipitation of AgCl when aqueous solu

Question

The balanced complete ionic equation for precipitation of AgCl when aqueous solution of LiCl and AgNO_3 are mixed is: Cl^- (aq) + Ag^+ (aq) rightarrow AgCl(l) Li^+ (aq) + Cl^- (aq) + Ag^+ (aq) + NO_3^- (aq) rightarrow AgCl(l) + Li^+ (aq) + NO_3^+ (aq) Ag^+ (aq) + Li^+ (aq) rightarrow Cl^- (aq) + NO_3^- (aq) Li^+ (aq) + NO_3^- (aq) rightarrow LiNO_3 (aq) The pressure of N_2 in a 15.0 L flask is 0.424 atm at 317.15 K. What mass of N_2 is in the flask? (R = 0.08206 L atm/mol middot K) 0.244 g 6.84 g 2.82 g 15.2 g 56.3 g What is the correct number of protons and neutrons in^Ba ? 56 protons and 56 neutrons 56 protons and 135 neutrons 56 protons and 79 neutrons 135 protons and 56 neutrons 135 protons and 79 neutrons

Explanation / Answer

18)   Li+(aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)   ---> AgCl (s) + Li+(aq) + NO3^ - (aq)

hence option B

19) V = 15 L ,   P = 0.424 atm . T = 317.15 K

PV = nRT

0.424 atm x 15 L = n x 0.08206 liter atm/molK x 317.15 K

n = 0.2444

Mass of N2 = moles of N2 x molar mass of N2

      = 0.2444 mol x 28 g/mol

       = 6.84 g

20) 135 Ba 56

atomic number = number of protons = 56

mass number = 135 = number of p + number of n

135 = 56 + n

n = 79

hence number of protons = 56 , number of neutrons = 79

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