Anabolic reactions may be characterized as
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Answer:
The answer is below
Explanation:
A normal model is represented as (μ, σ). Therefore for (1.5, 0.18), the mean (μ) = 1.5 and the standard deviation (σ) = 0.18
The z score shows by how many standard deviations the raw score is above or below the mean. It is given as:
a) For x < 1.35 s
From the normal distribution table, the percent of drivers have a reaction time less than 1.35 seconds = P(x < 1.35) = P(z < -0.83) = 0.2033 = 20.33%
b) For x > 1.9 s
From the normal distribution table, the percent of drivers have a reaction time greater than 1.9 seconds = P(x > 1.9) = P(z > 2.22) = 1 - P(z<2.22) = 1 - 0.9868 = 0.0132 = 1.32%
c) For x = 1.45
For x = 1.75
From the normal distribution table, P(1.45 < x < 1.75) = P(-0.28 < z < 1.39) = P(z < 1.39) - P(z< - 0.28) = 0.9177 - 0.3897 = 0.528 = 52.8%
d) A percentage of 10% corresponds to a z score of -1.28
e) P(z < z1) - P(z< -z1) = 60%
P(z < z1) - P(z< -z1) = 0.6
P(z < -z1) = 1 - P(z < z1)
P(z<z1) - (1 - P(z < z1)) = 0.6
2P(z<z1) - 1= 0.6
2P(z<z1) = 1.6
P(z<z1) = 0.8
From the z table, z1 = 0.85
The reaction time between 1.35 and 1.65 seconds
The mass sample of nicotine combusted = 2.625 mg (given)
Mass of produced = 7.1210 mg (given)
Mass of produced = 2.042 mg (given)
Molar mass of = 44 g/mol
Molar mass of = 18 g/mol
Percentage of Carbon = %
Percentage of hydrogen = %
Now, for percentage of nitrogen = %
Calculating the moles of each element:
- For
- For
- For
Dividing with the smallest mole value to calculate the molar ratio of each element:
Hence, the empirical formula for nicotine is .