How many milliliters of 0.132 M HCIO_4 solution are needed to neutralize 50.00 m

Question

How many milliliters of 0.132 M HCIO_4 solution are needed to neutralize 50.00 mL of 0.0789 M NaOH? A) 83.7 B) 0.0335 C) 0.0120 D) 0.521 E) 29.9 The net ionic equation for the reaction between aqueous solutions of HF and KOH is _____. A) H^+ + OH^- rightarrow H_2O B) HF + KOH rightarrow H_2O + K^+ + F^- C) H^+ + F^- + K^+ + OH^- rightarrow H_2 O + K^+ + F^- D) HF + OH^- rightarrow H_2O + F^- E) HF + K^+ + OH^- rightarrow H_2O + KF Which of the following are strong electrolytes? HCL HC_2H_3O_2 NH_3 KCI KCI A) 1, 2, 4 B) 1, 4 C) 1, 3, 4 D) 1, 2, 3, 4 E) 2, 4 Of the species below, only _____ is nm an electrolyte. A) NaCl B) Rb_2SO_4 C) Ar D) KOH E) HCI Which solution contains the largest number of moles of chloride ions? A) 30.00 mL of 0.100M CaCl_2 B) 750 mL of 0.50GM FeCl_3 C) 25.00 mL of 0.400M KCl D) 4.00 mL of 1.000M NaCl E) 10.0 mL of 0.500M BaCl_2 How many moles of K^+ are present in 343 mL of a 1.27 M solution of K_3PO_1? A) 0.145 B) 3.70 C) 1.31 D) 0.436 E) 11.1 The compound HCIO_4 is a _____. A) strong base B) weak base O strong acid D) weak acid E) binary compound Which of the following is not an exchange reaction? A) 2NaN_3(s) rightarrow 2Na(s) + 3N_2(g) B) H_2CO_3 (aq) + Ca(NO_3)_2 (aq) rightarrow 2HNO_3 (aq) + CaCO_3(s) C) NaC_2H_3O_2 (aq) + HNO_3(aq) rightarrow HC_2H_3O_2 (aq) + NaNO_3(aq) D) AgNO_3(aq) + HCl (aq) rightarrow AgCl(s) + HNO_3 (aq) E) Ba(C_2H_3O_2)_2 (aq) + Na_2SO_4(aq) rightarrow BaSO_4 (s) + 2NaC_2H_3O_2 (aq) A 31.5 mL aliquot of HNO_3 (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base to reach the endpoint of the titration. What was the concentration (M) of the acid? A) 0.0051 B) 1.02 C) 0.227 D) 0.0204 E) 0.0102

Explanation / Answer

Q22.

For the reaction:

NaOH + HClO4 --> NaClO4 + H2O

ratio is 1:1

1 mol of base = 1 mol of acid

so...

mol of base = MV = (0.0789*50) = 3.945*10^-3 mol

mol of acid = mol of base = 3.945*10^-3 mol

so...

for the acid

M = mol/V

V = mol/V = ( 3.945*10^-3)/(0.132) = 0.0298863 Liters

for mL --> 0.0298863*10^3 = 29.86 --> 29.9 mL

Q23.

For net ionic equation between acids, we must get rid of spectator ions:

HF(aq) + KOH(aq) --> H2O(l) + KF(aq)

split in ions

H+(aq) + F-(aq) + K+(aq) + OH-(aq) --> H2OI(l) + K+(aq) + F-(aq)

cancel common ions

H+(aq) + OH-(aq) --> H2OI(l)

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