29. ins 35 mL ofa o 1036 M NaoH is required to neutralize 250 ml. ara HCI soluti

Question

29. ins 35 mL ofa o 1036 M NaoH is required to neutralize 250 ml. ara HCI solution to phenolpbrhaleinend point, what is the mol larity of the HCT solution? a) 0.1036 M b00 2012 M e) 00636 M do 0,1272 M e none of these 20 Which molecules iv) HF e) ii ani d) ii and iv e) all are polar b) ii and iv 3L. What volume of 029s8 MHG will 123 g of Mgo a) 20.56 mL bu 18,84 ml ml do loss e) none of these 32. What is the molurkyora KMnow wourivn ir23 00mi is required to oxidize exactly 25.00 mL ofa 01015 MionllosadGresolwNon under acidic conditions? d) 0.1000M e) none of these

Explanation / Answer

29) NaOH + HCl -----------> NaCl + H2O

millimoles of NaOH = 15.35 x 0.1036 = 1.59

1.59 millimoles of HCl must be present.

1.59 = 25 x M

M = 0.0636

[HCl] = 0.0636 M

answer = option C = 0.0636 M

30) answer = option b = III and IV

HCl and HF are polar.

31) MgO + 2HCl -------------> MgCl2 + H2O

40.3 g MgO reacts with 2 moles HCl

0.1123 g MgO reacts with 0.1123 x 2 / 40.3 = 0.00557 moles HCl

moles = Molarity x volume in L

volume in L = moles / Molarity = 0.00557 / 0.2958 = 0.0188 L = 18.8 mL

answer = option b = 18.84 mL

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