Question 5 of 10
1 answer:
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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 question is incomplete, here is the complete question:
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
<u>Answer:</u> The percent yield of water in the reaction is 46.85 %.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaOH:</u>
Given mass of NaOH = 77.0 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
- <u>For sulfuric acid:</u>
Given mass of sulfuric acid = 72.6 g
Molar mass of sulfuric acid = 98 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So, 0.741 moles of sulfuric acid will react with = of NaOH
As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.
Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sulfuric acid produces 2 moles of water
So, 0.741 moles of sulfuric acid will produce = of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 1.482 moles
Putting values in equation 1, we get:
To calculate the percentage yield of water, we use the equation:
Experimental yield of water = 12.5 g
Theoretical yield of water = 26.68 g
Putting values in above equation, we get:
Hence, the percent yield of water in the reaction is 46.85 %.