Chemistry, 07.04.2020 21:13 naynay1997

PLZZZ HEL
1) Given the following balanced equation:
2Cs + F2 ---> 2CsF
- List each element on the reactant side and each element on the product side and assign the correct oxidation number to each.
- Identify which element is oxidized in the reaction and which is reduced.
- Identify the oxidizing agent in this reaction.

2) Balance the following redox equation using the oxidation-number-change method (you must use and demonstrate this method to earn credit). Show all of your work, describing each step.

Sn + HNO3 ---> SnO2 + NO2 + H2O

3) Explain two ways that voltaic cells and electrolytic cells are similar AND two ways that they differ.

Answers: 1

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Answers

Answer from: tinagibbs98

Explanation:

1). 2Cs + F2 ---> 2CsF

assigning the oxidation number to reactants and products of the above reaction:

Reactant = Cs + F2 (oxidation number of Cs = +1, oxidation number of F2 = 0)

Product = CsF ( Oxidation number of Cs in CsF is +2 and F is -1)

oxidized element in the reaction is one with increase in oxidation number

Oxidation number of Cs increases from +1 to +2

F is reduced because oxidation number is decreased from 0 to -1.

F is the oxidizing agent because it got reduced in the reaction.

2). Balanced equation:

Sn + HNO3 ---> SnO2 + NO2 + H2O

oxidation numbers 0 +1+5 -2 ⇔ -2+4-2 +4-2 +1-2

oxidation reaction= Sn ⇒ SnO2 + 4 e-

0 +4 -2

reduction reaction = HNO3 + e- ⇒ NO2

+1+5-2 +-2

combining the oxidation and reduction reaction equation, we will balance them by balancing the charges on them.

Sn + 2H2O ⇒ SnO2 + 4 e- +4 H+

HNO3 + e-+H+ ⇒ NO2 +H20

final reaction=

Sn + 4HNO3 SnO2 + 4 NO2 +2H20

0 +1+5-2 +4-2 +4-2

Charges on both the sides of reactant is

4 = 4

hence equation is balanced.

3 ) Voltaic cell and electrolyte cell difference is that voltaic cell converts chemical energy into electrical energy. Electrons are provided by redox reactions while in electrolyte cell electrical energy is converted to chemical energy and electron is given by external AC or DC battery.

Voltaic and electrolyte are similar as both these convert energy forms and require electron source for the reaction.

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.

Answer from: Quest

football

explanation:

because it is