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HCIO is titrated with a 0.273 (molarity) before any base is added? A) 0.723 B) 0

ID: 580781 • Letter: H

Question

HCIO is titrated with a 0.273 (molarity) before any base is added? A) 0.723 B) 0273 C)1.00-10-7 D) 2.81 10-13 E) 0.439 M KOH and 50.0 mL of o.125 M HCl is 8) 8) The pH of a solution prepared by mixing 50.0 ml of 0.125 B) 6.29 E) 0.00 D) 5.78 C) 8.11 9) A 25.0 ml sample of an HCl solution is titrated with a0.139 M NaOH solution. The equivalence C) 0.139 A) 7.00 M. point is reached with 15.4 mL of base. The concentration of HCl is E) 0.0856 A) 0.00214 B) 11.7 D) 0.267 10) Determine the Kup for magnesium hydroxide (Mg(OH)2) where the solubility of Mgo)2 is 1.41 10-4 M A) 2.7 10-12 B) 2.0 10-8 C) 1.110-11 D) 1.4 10-4 E) 3.9 × 10-8 12.0 PHo 10.0 soltution 8.0 in flask 6.0 Equivalence Point 4.0 2.0 5 10 15 20 25 30 35 40 45 mL of 0.115 M NaOH added to flask A 25.0 ml. sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. The titration curve above was obtained. The unknown compound is A) a strong acid B) a weak base C) a strong base D) a weak acid E) neither an acid nor a base he first law of thermodynamics can be given as A) for any spontaneous process, the entropy of the universe increases B) the entropy of a pure crystalline substance at absolute zero is zero D) ): grew/T at constant temperature E) AH'm =AH"(products)-JuHf (reactants)

Explanation / Answer

8)

we have:

Molarity of HCl = 0.125 M

Volume of HCl = 50 mL

Molarity of KOH = 0.125 M

Volume of KOH = 50 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.125 M * 50 mL = 6.25 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.125 M * 50 mL = 6.25 mmol

6.25 mmol of both will react to form neutral salt and water

So, pH will be 7.00

9)

we have the Balanced chemical equation as:

NaOH + HCl ---> NaCl + H2O

Here:

M(NaOH)=0.139 M

V(NaOH)=15.4 mL

V(HCl)=25.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HCl

1*M(NaOH)*V(NaOH) =1*M(HCl)*V(HCl)

1*0.139*15.4 = 1*M(HCl)*25.0

M(HCl) = 0.0856 M

Answer: E

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