PLEASE ANSWER ALL QUESTIONS THERE ARE FOUR DIFFERENT PROBLEMS 1.). A solution is
ID: 485684 • Letter: P
Question
PLEASE ANSWER ALL QUESTIONS THERE ARE FOUR DIFFERENT PROBLEMS
1.). A solution is made by dissolving 11.5 g of barium sulfide, BaS, in enough water to make exactly 250. mL of solution. Calculate the molarity of each species:
BaS mol/L
Ba2+ mol/L
S2- mol/L
2.).
Phenolphthalein
[NaOH] = 0.500 M
Set [NaOH] and add base in increments using the buttons above.
A beaker contains a 25 mL solution of an unknown monoprotic acid that reacts in a 1:1 stochiometric ratio with NaOH. Titrate the solution with NaOH to determine the concentration of the acid.
Perform a titration by setting the concentration of the NaOH solution and adding it to the acid solution using the different Add Base buttons.
The equivalence point of the titration is passed when the solution color changes.
The unknown sample can be titrated multiple times by pressing the Retitrate button and starting over.
3.). Enter the concentration of the unknown acid solution.The following thermochemical equation is for the reaction of hydrogen chloride(g) with ammonia(g) to form ammonium chloride(s).
HCl(g) + NH3(g)-------------->NH4Cl(s) DELTA H = -176 kJ
How many grams of HCl(g) would have to react to produce 35.4 kJ of energy?
4.) I got the fourth one, dont worry about that question! Thanks!
Explanation / Answer
molarity = W*1000/G.M.Wt* volume of solution in ml
= 11.5*1000/169.4*250
= 0.27M
BaS -----------------> Ba+2 + S-2
0.27M 0.27M 0.27M
2. NaOH + HA --------> NaA + H2O
no of moles of NaOH = molarity * volume in L
= 0.5*0.025 = 0.0125 moles
1 mole of NaOH react with 1 moles of HA
0.0125 M NaOH reat with 0.0125 M Of HCl
no of moles of HA = molarity * volume in L
0.0125 = molarity*0.025
molarity = 0.0125/0.025 = 0.5M
3. HCl(g) + NH3(g)-------------->NH4Cl(s) DELTA H = -176 kJ
1 mole of HCl react with NH3 to gives -176KJ
36.5g of HCl react with NH3 to gives -176KJ
36.5*35.4/176 = HCl react with NH3 to gives 35.4KJ
7.34g of HCl
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