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d1i1m1o1n [39]
2 years ago
6

The cutoff frequency for a certain

Chemistry
1 answer:
nikitadnepr [17]2 years ago
4 0

The cutoff frequency for a certain element is 1.22 x 1015 Hz this is its

work function in eV that is Work done = 5.05 eV.

<h3>What is supposed through a piece function?</h3>

Definition of labor function is the power this is wished for a particle to return back from the indoors of a medium and spoil via the surface —used particularly of the photoelectric and thermionic emission of electrons from metals paintings function.

  1. Work function = hf
  2. h= planks constant.
  3. f= frequency.
  4. W = 6.626 10^-34  1.22* 10^15.
  5. W = 8.08  10^-19 / ( 1.6 10^-19)
  6. W = 5.05 eV.

Read more about work:

brainly.com/question/1501489

#SPJ1

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Ferrous ammonium sulfate is what kind of test method.
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Two things

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2 years ago
Which best describe the tyndall effect
Svet_ta [14]

The answer would be A

6 0
3 years ago
In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After s
mart [117]

Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

0.3g Mg*\frac{16gO}{24.3gMg}= 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

0.2g O*\frac{1 mol O}{16g O}= 0.01mol O

0.3g Mg*\frac{1mol Mg}{24.3g Mg}= 0.01mol Mg

The moles of MgO can be obtained from:

0.5g MgO*\frac{1mol MgO}{40.3g MgO}= 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of  Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the  formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.

4 0
3 years ago
Nvermind i messed u. thats not what i meant to outtb ycgdgc8ydsaox agdydac bc7w
jasenka [17]

Answer:

same

Explanation:

yes, I agree 100% and you are correct *applaud*

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Calculate the number of moles of Al2O3 that are produced when 1.2 mol of FeO is used in the following reaction.
Assoli18 [71]

Answer:

D

Explanation:

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