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Sunny_sXe [5.5K]
2 years ago
6

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

Chemistry
1 answer:
son4ous [18]2 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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In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)
Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0
2 years ago
The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?
nordsb [41]
Mass of the gas m = 1.66 
The calculated temperature T = 273 + 20 = 293
 We have to calculate molar mass to determine the gas
 Molar Mass = mRT / PV
 M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
 M = 4043.76 / 101.3 = 39.92 g/mol
 So this gas has to be Argon Ar based on the molar mass.

7 0
1 year ago
Read 2 more answers
A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

V1 = Volume = 10L

V2= 2.4L

T1 = 0°C + 273 K = 273 K

T2 = 100°C + 273 K = 373 K

We have to find the pressure of the gas, by using the gas formula as,

$\frac{P 1 V 1}{T 1}=\frac{P 2 V 2}{T 2}

P2 can be found by rewriting the above expression as,

$P 2=\frac{P 1 \times V 1 \times T 2}{T 1 \times V 2}

Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

4 0
2 years ago
What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.
Mamont248 [21]

Answer:

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene

Explanation:

C_3H_6+HBr\rightarrow C_3H_7Br

Moles of propene = \frac{25.5 g}{39 g/mol}=0.6538 mol

According to reaction, 1 mole of propene gives 1 mole of propane.

Then 0.6538 moles of bromo-propane will give:

0.6538 mol\times 120 g/mol=78.46 g

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene.

5 0
2 years ago
Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or
givi [52]

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=68.9yrs

Putting values in above equation, we get:

k=\frac{0.693}{68.9}=0.0101yr^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.0101yr^{-1}

t = time taken for decay process = 206.7 yrs

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}

[A_o]=11.3g

Hence, the initial amount of Uranium-232 present is 11.3 grams.

4 0
2 years ago
Read 2 more answers
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