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2 years ago

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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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.

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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)

Sergio039 [100]

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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