Answer:
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In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis.
<span>Following are the investigative question that you can answer by doing this experiment.
</span>1) What information can be obtained from the colour of flame?
2) <span>State the relationship between wavelength, frequency, and energy?
</span><span>3) Can you identify the metal present in unknown sample provided?
4) How will you identify amount of metal present in sample solution?
5) </span><span>Why do different chemicals emit light of different colour?</span><span>
</span>
Answer: the speed at which products form
Explanation:
Rate of a reaction is defined as the speed at which a chemical reaction proceeds. It is often expressed in terms of the concentration of a reactant that is consumed in a unit time or the concentration of a product that is formed in a unit of time.
For a general reaction :
![Rate=-\frac{d[A]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D)
or ![Rate=+\frac{d[B]}{dt}](https://tex.z-dn.net/?f=Rate%3D%2B%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
where d[A] = change in concentration of reactant A
d[B] = change in concentration of product B
dt = time interval
D. <span>carbon and hydrogen</span>
Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
