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
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Answer:
6.9 (two sig figs)
Explanation:
2.375 + 4.5 = 6.875 = 6.9
When adding or subtracting, sig figs are determined by the least number of digits past the decimal point.
C. CaCl2 is the correct answer
Answer:
0.0400 g for the example given below.
Explanation:
pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.
- By definition, .
- NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: .
- From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: .
- Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: .
- We also know that . Take the antilog of both sides: .
- Solve for the mass of NaOH: .
Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain: