Potential energy is the energy stored in a chemical bond
While the normal gas flame can
only produce a “operating” to “light blue” type of flame, the Bunsen burner can
at least yield three types of flame. Consequently, the following: <span><span />
Operating flame
– which is yellow/orange in color, near 300° C. </span>
<span><span>·
</span>
Blue flame –
can be imperceptible under normal lighting conditions, near 500° C. The typically
used laboratory type of flame.</span>
<span><span>·
</span>Roaring-blue
flame – forms a triangular shaped in the center of the flame normally light
blue in color and interestingly, it’s a sound-producing flame. Heat is near to
700° C. </span>
Imagine with this three kinds
of flame produced and a Bunsen burner creates compared to a simple normal gas
flame. In sense, the roaring-blue flame proves evident as to why Bunsen burner
is hotter hence, the amount of heat it makes (700°C) it makes.
The correct answer is 0.16138
<u>Solution:</u>
Percentage transmission (%T) of the sample = 15.6
Therefore, Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069
<u>Use Beer’s law:</u>
A = ε*c*l where ε = molar absorptivity of the solution, c = concentration of the solution, and l = path length of the solution.
Given A = 0.8069 and l = 5.00 cm, we must have
<u>0.8069 = ε*c*(5.00 cm)</u>
ε*c = 0.16138 cm-1
For the next part, assume that ε*c remains constant and we have
A’ = ε*c*l’ where l = 1.00 cm.
Plug in values and write
A’ = (0.16138 cm-1)*(1.00 cm) = 0.16138
The ratio of the light intensity entering the sample to the light intensity exiting the sample at a particular wavelength is defined as the transmittance. Absorption and transmission are two related and different quantities used in spectroscopy. The main difference between absorption and transmission is that absorption measures how much incident light is absorbed as it travels through the material, while transmission measures how much light is transmitted.
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A) Since the plot 1/[AB] vs time gives straight line, the order of the reaction with respect to A is second order:
rate constant, K = slope = 5.5 x 10⁻² M⁻¹S⁻¹
b) Rate law : Rate = k[AB]²
c) half life period of the 2nd order is inversely proportional to the initial concentration of the reactants
t 1/2 =
.
t 1/2 =
d) k = 5.5 x 10⁻² M⁻¹s⁻¹
Initial concentration of AB, [A₀] = 0.250 M
concentration of AB after 75 s = [A]
k =
![\frac{1}{t} [ \frac{1}{[A]} - \frac{1}{[Ao]} ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bt%7D%20%5B%20%5Cfrac%7B1%7D%7B%5BA%5D%7D%20-%20%20%5Cfrac%7B1%7D%7B%5BAo%5D%7D%20%5D)
[A] = 0.123 M
Equation: AB → A + B
concentration of AB after 75 s = 0.123 M
Amount of AB dissociated = 0.25 - 0.123 = 0.127 M
concentration of [A] produced = concentration of [B] produced = Amount of AB reacted = 0.127 M
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
