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abruzzese [7]
3 years ago
8

Plz help on last question i asked

Chemistry
2 answers:
maxonik [38]3 years ago
8 0

Answer:

The total energy but the photons will give is 6.615 x 10^-14

Explanation:

klasskru [66]3 years ago
5 0

Answer:

The total energy but the photons will give is 6.615 x 10^-14

Explanation:

Formula that can be used in this question is E= hf or E= hc/ λ

This is the formula that can be used to calculate the energy that the photon is carrying where it is given as the Planck's Constant .

Now putting the values given in the equation that can calculate energy we get Planck's Constant is equal to 6.62 x 10^-34

E=(6.62 x 10^-34 x 2.998 x 10^8) / ( 3 x 10^-12 )

E=6.615 x 10^-14

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Which of the following elements have properties similar to those of astatine?
vekshin1
<h3><u>Answer;</u></h3>

F, Br, l

<h3><u>Explanation;</u></h3>
  • Astatine is a member of the halogen family, elements in Group 17 (VIIA) of the periodic table.
  • It is a highly radioactive element and it is the heaviest known halogen. Astatine is similar to the elements above it in Group 17 (VIIA) of the periodic table, especially iodine.
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3 0
3 years ago
Read 2 more answers
Can colloidal suspensions be separated out by filtration
balu736 [363]

Answer:

Colloidal can not be separated through filtration.

Suspension can be separated through filtration.

Explanation:

Colloidal:

Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

Suspension:

The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.

5 0
3 years ago
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how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze soluti
Tasya [4]

Answer: 2Liters

Explanation:

The expression used will be :

M_1V_1+M_2V_2=M_3V_3

where,

M_1 = concentration of first antifreeze= 60%

M_2 = concentration of second  antifreeze= 10%

V_1 = volume of first antifreeze = x L

V_2 = volume of second antifreeze = 8 L

M_3 = concentration of final antifreeze solution= 20%

V_3 = volume of final antifreeze = (x+8) L

Now put all the given values in the above law, we get the volume of  antifreeze added

60\times x+10\times 8=20\times (x+8)

x=2L

Therefore, the volume of 60% antifreeze solution that must be added is 2L

5 0
3 years ago
the density of aluminum is 2.70g/cm^3. a piece of aluminum foil has a volume of 54.0 cm^3. what is the mass of this piece of alu
Neporo4naja [7]
The  mass  of  aluminium  foil   is   calculated   as  follows
mass  =  density  x  volume
density =  2.70  g/cm^3
volume  54  cm^3
mass  of   aluminium  foil  is  therefore  =  2.70  g/cm^3  x  54  cm^3  =145.8  grams
cm^3  cancel   out  each   other
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What can you say about the amount of electrons found in the second ring
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Answer:

The second ring in an atom can only hold up to 8 electrons.

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