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Soloha48 [4]
3 years ago
10

Protein x has an absorptivity of 0.4 ml·mg-1 ·cm-1 at 280 nm. What is the absorbance at 280 nm of a 2.0 mg ·ml-1 solution of pro

tein x? (assume the light path is 1cm).
Chemistry
1 answer:
Evgen [1.6K]3 years ago
3 0

Absorbance measures the ability of the substance to absorb light at a specific wavelength.

Absorbance is also equal to the product of molar absorptivity, path length and molar concentration.

The mathematical expression is given as:

A= \epsilon l c       (1)

where, A = absorbance

\epsilon =  molar absorptivity

l = path length

c  = molar concentration.

The above formula is said to Beer's Law.

Absorptivity of protein x  = 0.4 mLmg^{-1}cm^{-1}

Path length = 1 cm

Molar concentration = 2.0 mg mL^{-1}

Put the values in formula (1)

Absorbance at 280 nm = 0.4 mL mg^{-1}cm^{-1}\times 1 cm \times 2.0 mg mL^{-1}

= 0.8

Thus, absorbance at 280 nm = 0.8

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What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?
Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

5 0
3 years ago
How do waves interact with matter ?
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Waves interact with matter in several ways. The interactions occur when waves pass from one medium to another. Besides bouncing back like an echo, waves may bend or spread out when they strike a new medium. These three ways that waves may interact with matter are called reflection, refraction, and diffraction.
6 0
3 years ago
Which of the following can be a
yanalaym [24]
The answer would be A
6 0
2 years ago
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca
Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

3 0
3 years ago
In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2
antiseptic1488 [7]

Answer:

The oxidizing agent is the MnO₄⁻

Explanation:

This is the redox reaction:

10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)

Let's determine the oxidation and the reduction.

I⁻ acts with -1 in oxidation state and changes to 0, at I₂.

All elements in ground state has 0 as oxidation state.

As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.

In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine

3 0
3 years ago
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