Answer:
Find the domain and the range of the following:
x y
3 2
5 7
1 4
9 2
3 7
Explanation:
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
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