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1 year ago

(LAB 9: SPECTROSCOPY: ABSORPTION & EMISSION)What are the four steps of the scientific method?

Chemistry

1 answer:

nata0808 [166]1 year ago

7 0

Answer

The four steps of the scientific method are:

1) Observation and description of a natural phenomenon of group of phenomena;

2) Formulation of a hypothesis to explain the phenomena;

3) Use of the hypothesis to predict the existence of other phenomena or to predict quantitatively the results of new observations, i.e to test the hypothesis if the hypothesis is true, find more evidence or find counter-evidence, if the hypothesis is false, create a new hypothesis or try again; and

4) Draw conclusions and repeat –the scientific method is never-ending, and no result is ever considered perfect.

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Answer:

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Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

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2 years ago

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Answer:

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Calculate the standard emf for the following reaction:

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In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
E = +1.47

Br(l) + 2e- = 2Br-
E = +1.065

We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V

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3 years ago