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2 years ago

What is the wavelength of a photon with an energy of 3.50 x 10^-19 J ?

Chemistry

1 answer:

Leokris [45]2 years ago

8 0

Answer:

λ = 5.68×10⁻⁷ m

Explanation:

Given data:

Energy of photon = 3.50 ×10⁻¹⁹ J

Wavelength of photon = ?

Solution:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = 3×10⁸ m/s

Now we will put the values in formula.

3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ

λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J

λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J

λ = 5.68×10⁻⁷ m

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An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?

juin [17]

Answer:

$pKa=3.58$

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

$pH=-log([H^+])$

$[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}$

And the percent ionization is:

$\% \ ionization=\frac{[H^+]}{[HA]}*100\%$

We compute the concentration of the acid, HA:

$[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\$

$[HA]=6.03x10^{-4}$

Thus, the Ka is:

$Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}$

So the pKa is:

$pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58$

Regards.

5 0

2 years ago

Questions

AlekseyPX

Answer:

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6 0

2 years ago

Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me

LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0

2 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

3 0

2 years ago

Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium

Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

5 0

2 years ago