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elena-14-01-66 [18.8K]
2 years ago
10

Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. The energy of one photon of this

light is____________.
A) 1.04*10^-22

B)1.4*10^-31

C) 2.64*10^18

D) 3.79*10^-28

E) 3.79*10^-19
Chemistry
1 answer:
Sedbober [7]2 years ago
3 0

<u>Answer:</u> The energy of one photon of the given light is 3.79\times 10^{-19}J

<u>Explanation:</u>

To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

\lambda = wavelength of light = 525nm=5.25\times 10^{-7}m        (Conversion factor:  1m=10^9nm  )

h = Planck's constant = 6.625\times 10^{-34}J.s

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

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{5.25\times 10^{-7}mm}\\\\E=3.79\times 10^{-19}J

Hence, the energy of one photon of the given light is 3.79\times 10^{-19}J

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vovikov84 [41]

The empirical formula is the simplest formula of a chemical compound.

To find the empirical formula, we take the following steps;

  • Divide the percentage by mass of each element by its relative atomic mass.
  • Divide the quotient of each by the lowest value obtained instep 1 above
  • Write the result of step 2 above as the subscript following each atom.

1) O - 88.10/16,      H - 11.190/1

  O - 5.5,               H - 11.19

  O - 5.5/5.5,        H - 11.19/5.5

  O -  1,                 H - 2

Empirical formula = OH2

2) C - 41.368/12  H - 8.101/1,   N - 32.162/14,   O - 18.369/16

   C - 3,               H - 8,           N - 2,                  O - 1

   C - 3/1,            H - 8/1          N - 2/1                 O - 1/1

    C - 3,             H - 8,           N - 2,                   O - 1

Empirical formula = C3H8N2O

To obtain the molecular formula where n = number of atoms of each element;

Molecular weight = 174.204 g/mol

[ 3(12) + 8(1) + 2(14) + 16]n = 174

n= 174/88

n = 2 (to the nearest whole number)

Hence, we have;

[C3H8N2O]2

The molecular formula is C6H16N4O2

3)  C - 19.999/12,  H - 6.713/1,   N - 46.646/14,   O - 26.641/16

    C - 2,                H - 7,            N -  3,                 O - 2

    C - 2/2,            H - 7/2,         N -   3/2,             O - 2/2

    C - 1,                H - 4,            N -  2,                  O - 1

Empirical formula - CH4N2O

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3 years ago
Write a balanced half-reaction for the reduction of permanganate ion mno−4 to solid manganese dioxide mno2 in acidic aqueous sol
Svetlanka [38]

The half-reaction includes either the reduction or the oxidation reaction of the redox reactions. In acidic solution permanganate ion will react with hydrogen ion to yield manganese ion and water.

<h3>What are Redox reactions?</h3>

Redox or oxidation-reduction reactions are the chemical reactions in which the oxidation and the reduction of the chemical species occur simultaneously.

Permanganate (VII) ion is a strong oxidizing agent and gets easily reduced to manganese ion in presence of the hydrogen ion in an acidic solution.

The balanced half-reaction for reduction is shown as,

\rm MnO_{4}^{-} \; (aq)+ 8H^{+} \; (aq)+ 5e^{-} \rightarrow Mn^{2+} \; (aq)+ 4H_{2}O \; (l)

Learn more about reduction reactions here:

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2 years ago
GoOd MoRnInG <br> EvErY OnE <br> BlA BlA BlA BlA
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Answer:

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Explanation:

4 0
3 years ago
Read 2 more answers
What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
what are simplified definitions of the following words"Synthesis, Decomposition, Single Replacement, or Double Replacement"?
faust18 [17]

Answer:

The Four Major Types of Reactions

Name General Reaction Pattern

Combination or synthesis A + B ----> AB

Decomposition AB ----> A + B

Substitution or Single Replacement A + BC ----> B + AC

Metathesis or Double Displacement AB + CD ----> AD + CB

Combination or Synthesis Reactions Two or more reactants unite to form a single product.

S + O2 ---------> SO2

sulphur oxygen sulphur dioxide

2 S + 3 O2 ---------> 2 SO3

sulphur oxygen sulphur trioxide

2 Fe + O2 ---------> 2 FeO

iron oxygen iron (II) oxide

Decomposition Reactions A single reactant is decomposed or broken down into two or more

products.

CaCO3 ----------> CaO + CO2

calcium carbonate calcium oxide carbon dioxide

2 H2O -----------> 2 H2 + O2

water hydrogen oxygen

2 KClO3 -----------> 2 KCl + 3 O2

potassium chlorate potassium chloride oxygen

Substitution or Single Replacement Reactions A single free element replaces or is substituted for one of the elements in a compound. The free element is more reactive than the one its replaces.

Zn + 2 HCl ----------> H2 + ZnCl2

zinc hydrochloric acid hydrogen zinc chloride

Cu + 2 AgNO3 -----------> 2 Ag + Cu(NO3)2

copper silver nitrate silver copper (II) nitrate

H2 + 2 AgNO3 -----------> 2 Ag + 2 HNO3

hydrogen silver nitrate silver nitric acid

2 Na + 2 H2O -----------> 2 NaOH + H2

sodium water sodium hydroxide hydrogen

Metathesis or Double Displacement Reactions This reaction type can be viewed as an "exchange of partners." For ionic compounds, the positive ion in the first compound combines with the negative ion in the second compound, and the positive ion in the second compound combines with the negative ion in the first compound.

HCl + NaOH -----------> NaCl + HOH

hydrochloric sodium sodium water

acid hydroxide chloride

BaCl2 + 2 AgNO3 ----------> 2 AgCl + Ba(NO3)2

barium silver silver barium

chloride nitrate chloride nitrate

(precipitate)

CaCO3 + 2 HCl -----------> CaCl2 + H2CO3

calcium hydrochloric calcium carbonic

carbonate acid chloride acid

Explanation:

The vast number of chemical reactions can be classified in any number of ways. Under one scheme they can be categorized either as oxidation-reduction (electron transfer) reactions or non-oxidation-reduction reactions. Another completely different but common classification scheme recognizes

8 0
2 years ago
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