Chap 4 moles of silver 3. 43.8 mL of 1.49 M nitric acid is added to 42.7 mL of s

Question

Chap 4

moles of silver

3.43.8 mL of 1.49 M nitric acid is added to 42.7 mL of sodium hydroxide, and the resulting solution is found to be acidic.

24.6 mL of 0.927 M barium hydroxide is required to reach neutrality.

What is the molarity of the original sodium hydroxide solution?

Chap 6

2. a.Calculate the energy of an electron in the n = 3 level of a hydrogen atom.   Energy = ______ Joules

The reaction consumes _________ moles of silver nitrate The reaction produces _________ moles of copper(II) nitrate and ________

moles of silver

Explanation / Answer

1. For the given reaction,

we would multiply moles of H2SO4 with 2 to get moles of water formed.

2. If 3 moles of Cu has reacted then,

We would consume = 6 moles of AgNO3

we would produce = 3 moles of Cu(NO3)2 and,

                produce = 6 moles of Ag

3. moles of HNO3 present = 1.49 M x 0.0438 L = 0.065 mol

moles of HNO3 reacted with Ba(OH)2 added = 2 x 0.927 M x 0.0246 L = 0.045 mol

moles NaOH present = moles HNO3 reacting with NaOH = 0.065 - 0.045 = 0.02 mols

Molarity of NaOH = 0.02 mol/0.0427 L = 0.47 M

Chap 6.

1. Matching the responses

1) The emission line with shortest wavelength = C

2) Absorption line with longest wavelength = B

3) emission line with highest energy = C

4) absorption line with lowest energy : B

5) emisiion line with highest frequency = C

6) line corresponding to ionization energy of hydrogen = A

Energy is directly proportional to frequency and inversely related to wavelength

2. a. When n = 3,

Energy = 2.18 x 10^-18 x 1/3^2 = 2.42 x 10^-19 J

b. Energy for n = 6 to n = 8

dE = -2.18 x 10^-18(1/8^2 - 1/6^2) = 2.64 x 10^-20 J

This is an absorption (A) process.

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