1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dem82 [27]
3 years ago
7

Calculate the wavelength and frequency of an emitted photon of gamma radiation that has energy of 3.02 × 1011 J/mol.

Chemistry
1 answer:
maks197457 [2]3 years ago
6 0

The wavelength is 4.0 * 10^-13 m while the frequency is 7.5 * 10^20 s−1

Using the formula;

E =hf

Where

E = Energy of the photon

f = Frequency of photon

h = Plank's constant = 6.6 * 10^-34 Js

To obtain the energy in Joules;

E = 3.02 × 10^11 J/mol/6.02 * 10^23 mol

= 5 * 10^-13 J

f = E/h

f = 5 * 10^-13 J/6.6 * 10^-34 Js

f = 7.5 * 10^20 s−1

f = c/λ

c = speed of light = 3 * 10^8 m/s

λ = c/f

λ =  3 * 10^8 m/s/7.5 * 10^20 s−1

λ =  4.0 * 10^-13 m

The wavelength is 4.0 * 10^-13 m while the frequency is 7.5 * 10^20 s−1

Learn more: brainly.com/question/2393994

You might be interested in
____________ rays show areas where large quantities of energy are released.
sladkih [1.3K]

Answer:

Gamma

Explanation:

3 0
2 years ago
Please help I need a 100
svlad2 [7]

your answer would be C.

6 0
2 years ago
Read 2 more answers
A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde
xeze [42]

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

6 0
3 years ago
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silv
Flauer [41]

Answer:

1.1 × 10⁻⁴ M

Explanation:

Let's consider the following double displacement reaction.

CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)

We can establish the following relations:

  • The molar mass of AgCl is 143.32 g/mol.
  • The molar ratio of AgCl to CuCl₂ is 2:1

The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

7.7 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molCuCl_{2}}{2molAgCl} =2.7 \times 10^{-5}molCuCl_{2}

The molarity of CuCl₂ is:

M=\frac{2.7 \times 10^{-5}molCuCl_{2}}{0.250L} =1.1\times 10^{-4} M

7 0
2 years ago
Show me the periodic table
yawa3891 [41]
I think this is what you wanted, so good luck!

8 0
2 years ago
Other questions:
  • At that volume is measured to be 755 mm of Hg. If the lungs are compressed to a newA healthy male adult has a lung capacity arou
    12·1 answer
  • The red spheres represent oxygen atoms, and the blue spheres represent hydrogen atoms. Is this substance a compound?
    8·2 answers
  • A + B → C Select the rate law for the reaction above using the following information: Holding the concentration of A constant an
    15·1 answer
  • When a chemical reaction releases energy it is called what
    7·2 answers
  • What is a comet? tell in a short and easy answer​
    14·1 answer
  • Janice is given a mixture of alcohol and water. The teacher tells her that she can use temperature to separate these compounds.
    13·2 answers
  • Which matter exists in the liquid state at room temperature? A. lead B. nebula C. apple juice D. carbon monoxide
    14·1 answer
  • Vaccanation against the flu should not be required
    6·1 answer
  • What is the concentration of lithium ions in 0.400 M Li₂HPO₄?
    7·1 answer
  • To completely neutralise 200cm3 of 0.5mol/dm3 sodium hydroxide (NaOH), a student adds 100cm3 of 0.5mol/dm3 sulfuric acid (H2SO4)
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!