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djverab [1.8K]
2 years ago
11

Calculate the pH of an aqueous solution that is 0.020 M in HCl(aq) at 25°C.

Chemistry
1 answer:
sdas [7]2 years ago
7 0
Hope this helps you!

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It’s extremely bad quality I really can’t read it
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3 years ago
6Na(s) + N2(g) --> 2Na3N(s)
Ksivusya [100]

Answer:

 80.0 g Na and 20.0 g N2.

Explanation:

This means the limiting reactant determines the maximum mass of the product formed.

7 0
1 year ago
When an electron moves from a lower energy level to a higher energy level, the atom:
Ede4ka [16]

Answer:

A. absorbs light

Explanation:

For an electron to move from a lower energy level to a higher energy level, energy needs to be added to the atom.

In most cases this required energy will be light: The electron of the atom will absorb the energy of the light and move to a higher energy level.

When said electron comes back to the lower energy level, the atom will conversely give off light.

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2 years ago
How does hydrogen bonding affect water as it becomes colder and eventually freezes? (1 point)
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2 years ago
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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