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Anestetic [448]
3 years ago
8

Night visin cameras are sensitive to energies around 2.21 x 10-19 J. What wavelength of electromagnetic radiation do they use?

Chemistry
1 answer:
defon3 years ago
6 0

Answer:

8.99×10^-7m

Explanation:

The wavelength can be calculated using the expression below

E=hcλ

Where E= energy= 2.21 x 10^-19 J.

C= speed of light= 3x10^8 m/s

h= planks constant= 6.626 × 10^-34 m2 kg / s

E=hcλ

λ= E/(hc)

Substitute for the values

λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )

= 8.99×10^-7m

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A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

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Explanation:

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Convert 6.35 grams of aluminum sulfate to moles​
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Answer:

There are 0.0186 moles of formula units in 6.35 grams of aluminum sulfate \rm Al_2(SO_4)_3.

Explanation:

What's the empirical formula of aluminum sulfate?

Sulfate is an anion with a charge of -2 per ion. When sulfate ions are bonded to metals, the compound is likely ionic.

Aluminum is a group III metal. Its ions tend to carry a charge of +3 per ion.

The empirical formula of an ionic compound shall balance the charge on ions with as few ions as possible.

The least common multiple of 2 and 3 is 6. That is:

  • Three sulfate ions \rm {SO_4}^{2-} will give a charge of -6.
  • Two aluminum ions \rm Al^{3+} will give a charge of +6.

Pairing three \rm {SO_4}^{2-} ions with two \rm Al^{3+} will balance the charge. Hence the empirical formula: \rm Al_2(SO_4)_3.

What's the mass of one mole of aluminum sulfate? In other words, what's the formula mass of \rm Al_2(SO_4)_3?

Refer to a modern periodic table for relative atomic mass data:

  • Al: 26.982;
  • S: 32.06;
  • O: 15.999.

There are

  • two Al,
  • three S, and
  • twelve O

in one formula unit of \rm Al_2(SO_4)_3.

Hence the formula mass of \rm Al_2(SO_4)_3:

\underbrace{2\times 26.982}_{\rm Al} + \underbrace{3\times 32.06}_{\rm S} + \underbrace{12\times 15.999}_{\rm O} = \rm 342.132\;g\cdot mol^{-1}.

How many moles of formula units in 6.35 grams of \rm Al_2(SO_4)_3?

\displaystyle n = \frac{m}{M} = \rm \frac{6.35\;g}{342.132\;g\cdot mol^{-1}} = 0.0186\;mol.

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It changes from a liquid to a solid.
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4 years ago
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