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Sunny_sXe [5.5K]
3 years ago
6

Chromium (III) forms a complex with diphenylcarbazide whose molar absorptivity is 4.17x104 at 540 nm.

Chemistry
1 answer:
son4ous [18]3 years ago
5 0

Answer:

3.00 cm

Explanation:

The absorbance can be expressed using <em>Beer-Lambert's law</em>:

A = ε*b*c

Where ε is a constant for each compound, b is the optical path, and c is the molar concentration of the compound.

Now we <u>match the absorbance values for both solutions</u>, because we want the absorbance value to be the same for both solutions:

A = ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

And <u>solve for b:</u>

 ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

1.00 cm * 7.68x10⁻⁶M = b * 2.56x10⁻⁶ M

b = 3.00 cm

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3 years ago
Arrange the following bonds in order of increasing ionic character. assign number 1 as the highest and 6 as the lowest. a. carbo
SIZIF [17.4K]
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Explanation:
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5 0
2 years ago
While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake
ale4655 [162]

Answer:

11552.45 years

Explanation:

Given that:

Half life = 5730 years

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{5730}\ years^{-1}

The rate constant, k = 0.00012 years⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.00012 years⁻¹

Initial concentration [A_0] = 160.0 counts/min

Final concentration [A_t] = 40.0 counts/min

Time = ?

Applying in the above equation, we get that:-

40.0=160.0e^{-0.00012\times t}

e^{-0.00012t}=\frac{1}{4}

-0.00012t=\ln \left(\frac{1}{4}\right)

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7 0
3 years ago
How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
7 0
3 years ago
What is the number of electrons that can be held in the first orbit closest to the nucleus
Juliette [100K]
The first shell can hold up to 2 electrons, the second shell can hold up to 8 (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on.
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3 years ago
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