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zubka84 [21]
2 years ago
12

A student prepared an unknown sample by making a dilute solution of the unknown sample. The dilute sample was prepared by adding

5.0 mL of the unknown sample to a 25.0 mL flask. What is the concentration(mM) of the diluted unknown sample that has an absorbance of 0.270
Chemistry
1 answer:
telo118 [61]2 years ago
4 0

Answer:

0.27 mM

Explanation:

According to the law of Lambert Beer:

A = C × E × L

Where:

A = Absorbance

C = Concentration

E = molar absorptivity

L = Step Length

If me make C the subject of the formula, we have = A / E × L

We know Absorbance and we can assume L = 1 cm, but being an unknown substance, we must know the molar Absorptivity of it, therefore, only having that value we could calculate the concentration of the mixture using the previous equation.

Then we could use the dilution ratio:

Cc * Vc = Cd * Vd

from the above formula, to fin the  concentrated Concentration, we have:

Cc = Cd * Vd / Vc

we then replace the known values.

Vd = Diluted volume 25 mL

Vc = Volume concentrated 5 mL

and we would not know the diluted concentration (Cd), which is what we had to calculate in the first section. Substituting it in the previous equation, we can obtain the initial concentration.

Assuming a molar absorptivity of 5000 M-1 * cm -1 we would have:

Cd = 0.270 / 5000 * 1

= 5.4 x 10 ^ -5 M * (1000 mM / 1 M)

= 0.054 mM

and initial concentration:

Cc = 0.054 mM * (25 mL / 5 mL)

= 0.27 mM.

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Give four different chemical meanings for the word element and an example of each
Naddika [18.5K]

Answer ;

-An element is a substance containing only one type of atom, for example; H2 or 02  (consisting of atoms that all have the same number of protons).

-Microscopic, single atom of that element

-Macroscopic, sample of that element large enough to weigh on a balance

- A substance that cannot be broken down chemically; e.g; sodium metal,

Explanation;

-An element is a substance whose atoms all have the same number of protons: another way of saying this is that all of a particular element's atoms have the same atomic number. Elements are chemically the simplest substances and hence cannot be broken down using chemical reactions.

-An element is uniquely determined by the number of protons in the nuclei of its atoms.

3 0
2 years ago
What is the mass, in g, of 0.500 mol of 1,2-dibromoethane, CH2BrCH2Br?
inna [77]
Molar mass <span>CH2BrCH2Br = 188.0 g/mol

1 mole ---------- 188.0 g
</span>0.500 moles ----- ?

mass = 0.500 * 188.0 / 1

= 94.0 g

Answer C

hope this helps!
4 0
2 years ago
Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,
vfiekz [6]

Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0
2 years ago
The differential cross section for scattering 6.5-MeV alpha particles at 120° off a silver nucleus is about 0.5 barns/sr. If a t
Burka [1]

Answer:

Nsc=30

Explanation:

The solid angle subtended by the counter is

d=0.1mm2/(1cm)2=(10)3 sr

d=(10)10x10.5gcm3x10-4cm/108x1.6610x10-24gx0.510-24cm2x10-3=30

8 0
2 years ago
When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po
SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide (C_{7}H_{7}NO) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is 2.7^{o}C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH_{4}Cl) are dissolved in the same mass of X, the freezing point of the solution is 9.9^{o}C lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

    Moles of benzamide = \frac{mass}{\text{Molar mass of benzamide}}

                    = \frac{70.4 g}{121.14 g/mol}

                    = 0.58 mol

Now, we will calculate the molality as follows.

     Molality = \frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}

                   = \frac{0.58 mol}{0.85 kg}

                   = 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

      dT = i \times K_{f} \times m,

where,      dT = change in freezing point = 2.7^{o}C

                  i = van't Hoff factor = 1 for non dissociable solutes

      K_{f} = freezing point constant of solvent

                m = 0.6837

Therefore, putting the given values into the above formula as follows.

             dT = i \times K_{f} \times m,

            2.7^{o}C = 1 \times K_{f} \times 0.6837 m

            K_{f} = 3.949 C/m

Now, we use this K_{f} value for calculating i for NH_{4}Cl

So, moles of ammonium chloride are calculated as follows.

 Moles of NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}

                            = 1.316 mol

Hence, calculate the molality as follows.

    Molality = \frac{1.316 mol}{0.85 kg}

                  = 1.5484

It is given that value of change in temperature (dT) = 9.9^{o}C. Thus, calculate the value of Van't Hoff factor as follows.

              dT = i \times K_{f} \times m

   9.9^{o}C = i \times 3.949 C/m \times 1.5484 m

                     i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0
2 years ago
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