Question

A 12.9 mL sample of HNO3 was diluted to a volume of 104.11 mL. Then, 25 mL of that diluted solution were needed to neutralize 46.2 mL of 0.50 M KOH. What was the concentration of the original nitric acid?

Answer #1

A 12.8 mL sample of HNO3 was diluted to a volume of 100.52 mL.
Then, 26 mL of that diluted solution were needed to neutralize 46.0
mL of 0.47 M KOH. What was the concentration of the original nitric
acid?

What volume of H2 gas (in liters), measured at 27 °C and 737
torr, can be obtained by reacting 4.95 g of zinc metal with 121 mL
of 0.305 M HCl? The equation is______ L H2(g) A 0.228 g sample of
impure NaOH requires 17.1 mL of 0.2370 M HCl for neutralization.
What is the percent of NaOH in the sample, by weight?_____% NaOH In
an acid-base titration, 20.54 mL of H2SO4 were used to neutralize
33.96 mL of 0.322...

73.0 mL of a 1.60 M solution is diluted to a total volume of
288 mL. A 144 mL portion of that solution is diluted by adding 129
mL of water. What is the final concentration? Assume the volumes
are additive.
Determine the concentration of each of the individual ions in
a 0.350 M K2SO4 solution.
A 0.110 L sample of an unknown HNO3 solution required 51.1 mL
of 0.200 M Ba(OH)2 for complete neutralization. What was the
concentration of...

A sample 25.0 mL of an NaCl solution is diluted to a final
volume of 125.0 mL of 0.200 M NaCl.What was concentration of the
original NaCl solution?How many moles of solute are pipetted from
the original solution?How many moles of solute are in the dilute
solution?
How many moles of product, KHCO3 will be produced
when1.0 mol of K2O reacts with 1.0 mol H2O
and 1.0 mol of CO2?K2O + H2O + 2
CO2 à 2 KHCO3
h) A...

a) What volume of a 6.40 M stock solution do
you need to prepare 500. mL of a
0.0347 M solution of HNO3?
b) The absorbance of a cationic iron(II) sample solution was
measured in a spectrophotometer, but the instrument returned an
error because the absorbance was too high. The sample was then
diluted by using a pipette to take 100.0 μL of the sample
and injecting it into a cuvette already containing 2.00 mL of water
(total volume is 2.00 mL +...

The molarity of an aqueous solution of potassium
hydroxide ( KOH ) is determined by
titration against a 0.171 M nitric
acid ( HNO3 ) solution.
If 25.0 mL of the base are required to neutralize
10.4 mL of nitric acid , what is
the molarity of the potassium hydroxide
solution?

A) What volume of 0.150 M HClO4 solution is needed to neutralize
59.00 mL of 8.80×10−2 M NaOH? B) What volume of 0.130 M HCl is
needed to neutralize 2.87 g of Mg(OH)2? C) If 26.4 mL of AgNO3 is
needed to precipitate all the Cl− ions in a 0.755-mg sample of KCl
(forming AgCl), what is the molarity of the AgNO3 solution? D) If
45.8 mL of 0.112 M HCl solution is needed to neutralize a solution
of KOH,...

Part A
What volume of 0.105 M HClO4 solution is needed to
neutralize 55.00 mL of 9.00×10−2M NaOH?
Part B
What volume of 0.120 M HCl is needed to neutralize 2.70
g of Mg(OH)2?
Part C
If 25.6 mL of AgNO3 is needed to precipitate all the Cl− ions in
a 0.770-mg sample of KCl (forming AgCl), what is the molarity of
the AgNO3 solution?
Part D
If 45.7 mL of 0.102 M HCl solution is needed to
neutralize a...

A
25.0 mL sample of solution of unkown acid, HX, is titrated with
2.87 M KOH. If 59.9 mL of KOH was required to neutralize the
sample, find the molaritu of fhe original HX solution.

A volume of 70.0 mL of a 0.860 M HNO3 solution is
titrated with 0.450 M KOH. Calculate the volume of KOH
required to reach the equivalence point.

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