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Vikentia [17]
3 years ago
14

Can you tell the difference between an element and compound by measuring mass and volume?

Chemistry
1 answer:
vivado [14]3 years ago
8 0
Yes, you can!




Hope this helped, and brainliest much needed and appreciated! Have a wonderful day!:)
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Which models of the atom does the experimental evidence from Bohr's hydrogen experiment support? Explain why these models are co
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Answer:

Rutherfords

Explanation:

The model of the atom supported by Bohr's hydrogen experiment is the Rutherford's model of the atom.

Rutherford through his experiment on gold foil suggested the atomic model of the atom. The model posits that an atom has a small positively charged center(nucleus) where nearly all the mass is concentrated.

  • Surrounding the nucleus is the large space containing electrons.
  • In the Bohr's model of the atom, he suggested that the extranuclear space of the atom is made up of electrons in specific spherical orbits around the nucleus.
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In balancing an equation, we change the __________ to make the number of atoms on each side of the equation balance.
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If you are talking about chemistry, it's coefficient!
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A chemical bond between two atoms results from a simultaneous
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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
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