Chemistry, 29.03.2020 02:59 elmoreroman11

How many moles of ions are released when 0.27 mole of cobalt chloride is dissolved in water?

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Answer from: Deascry

0.81 moles

Explanation:

cobalt chloride when dissolved in water dissociate into cobalt and chloride ions as shown;

CoCl₂ (s) → [Co²⁺](aq) + 2[Cl⁻] (aq)

The mole ratio between CoCl₂ and [Co²⁺] is 1 ; 1 while the mole ratio between CoCl₂ and [Cl⁻] is 1 : 2

Therefore; 0.27 moles of CoCl₂ will form;

(1 * 0.27) = 0.27 moles of [Co²⁺] ions

(2 * 0.27) = 0.54 moles of [Cl⁻] ions

Total moles of dissociated ions are;

0.27 + 0.54

= 0.81 moles

Answer from: Quest

the answer is c. nacl

Answer from: Quest

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.

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