Chemistry, 07.03.2020 20:08, krishawnnn

A 50.0 mL sample of KCl requires 22.40 mL of 0.0229 M Pb(NO3)2 in order to completely titrate it. What is the Molarity of the KCl sample?

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answered: tali2561

0.02052M

Explanation:

First, we need to write a balanced equation for the reaction. This is illustrated below:

2KCl + Pb(NO3)2 → 2KNO3 + PbCl2

The following were obtained from the question:

Molarity of Pb(NO3)2 = 0.0229M

Volume of Pb(NO3)2 = 22.40 = 22.4/1000 = 0.0224L

Number of mole of Pb(NO3)2 =?

Recall:

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.0229x0.0224

Mole of Pb(NO3)2 = 5.13x10^-4mole

From the equation,

1mole of Pb(NO3)2 required 2moles KCl.

Therefore, 5.13x10^-4mole of Pb(NO3)2 will require = 5.13x10^-4x2 = 1.026x10^-3mole of KCl.

Now we can use this amount (i.e 1.026x10^-3mole) to find the molarity of KCl. This is illustrated below:

Mole of KCl = 1.026x10^-3mole

Volume of KCl = 500mL = 50/1000 = 0.05L

Molarity =?

Molarity = mole /Volume

Molarity of KCl = 1.026x10^-3/0.05

Molarity of KCl = 0.02052M

answered: Guest

the balanced equation for reaction of ca metal with o2 to produce cao is as below

2 ca+ o2→ 2 cao2

explanation

the reaction is balanced since the number of atoms in reactant side is equal to number of atoms in product side. for example they are 2 atom of ca in reactant side and 2 atoms of ca in the product side.

the process is a meet the requirement of redox reaction because : calcium is oxidized from oxidation state 0 to 2+ while oxygen is reduced from oxidation state 0 to 2-

answered: Guest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.