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wel
3 years ago
10

The following data were obtained for a complex of nickel at 575 nm in a 1.00 cm cell. Using Microsoft Excel or some other graphi

ng program, construct a calibration plot and accurately calculate the molar absorptivity (ε) for the nickel-complex at 575 nm.
Standard # Nickel Complex Concentration (M) Absorbance
1 0.01 0.058
2 0.02 0.109
3 0.04 0.220
4 0.06 0.328

A. 0.184 M-1cm-1

B. 0.002 M-1cm-1

C. 5.425 M-1cm-1

D. Not Enough Information to Determine

Chemistry
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

Check the explanation

Explanation:

It is basically known that the change in complex compositions of metal play a vital role in the conduction of molecular materials, which display atypical magnetic properties and conducting which also finds applicability in material chemistry, supramolecular and biochemistry, The electrochemical conduct of the nickel Ni(II) complexes was investigated in DMSO by cyclic voltammetry (CV), rotating coulometry and disc electrode (RDE).

The solution to the question above can be seen in the attached image below

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Name some acids and in what and how we used them
Ne4ueva [31]

Answer:

Acid Uses

Organic acids

Citric acid As a preservative for food As a flavouring agent

Ascorbic acid (also called vitamin C) In the treatment of bone marrow and scurvy diseases

Acetic acid Added to pickles to make them sour

Explanation:

7 0
2 years ago
. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
Tom [10]

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength = 435.8\times 10^{-9}m

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

3 0
3 years ago
What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi
Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL

3 0
3 years ago
In laboratory experiment, a NOVDEC Student was
Ede4ka [16]

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

<em>The volume of the glucose solution to be prepared</em> = 500 cm^3

<em>Molarity of the glucose solution to be prepared</em> = 1 M

i. Molar mass of glucose (C_1_2H_6O_6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii.<em> mole = molarity x volume</em>. Hence;

amount (in moles) of the glucose solution to be prepared

                 = 1 x 500/1000 = 0.5 mole

3 0
3 years ago
____ is the transfer of thermal energy by the movement of particles from one part of the material to another
REY [17]

Answer:

convection

Explanation:

4 0
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