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Digiron [165]
3 years ago
5

Calculate the percentage by mass of the indicated element in the following compounds. Hydrogen in ascorbic acid, HC6H7O6, also k

nown as vitamin C. Express your answer using two significant figures
Chemistry
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer: The mass percent of hydrogen in ascorbic acid is 4.5 %

Explanation:

In C_6H_8O_6, there are 6 carbon atoms, 8 hydrogen atoms and 6 oxygen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Molar mass of ascorbic acid}}\times 100

Mass of hydrogen = 8\times 1g/mol=8g

Molar Mass of ascorbic acid =6\times 12g/mol+8\times 1g/mol+6\times 16g/mol=176g

Putting values in above equation, we get:

\text{Mass percent of hydrogen}=\frac{8g}{176}\times 100=4.5\%

Hence, the mass percent of of hydrogen in ascorbic acid is 4.5 %.

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Avoiding an accident when driving can depend on reaction time. That time, measured in seconds from the moment the driver see dan
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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:

z=\frac{x-\mu}{\sigma}

a) For x < 1.35 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.35-1.5}{0.18}=-0.83

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

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.9-1.5}{0.18}=2.22

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

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.45-1.5}{0.18}=-0.28

For x = 1.75

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.75-1.5}{0.18}=1.39

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

z=\frac{x-\mu}{\sigma}\\\\-1.28=\frac{x-1.5}{0.18}\\\\x-1.5=-0.2034\\\\x=1.27

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

0.85=\frac{x-1.5}{0.18}and-0.85=\frac{x -1.5}{0.18}  \\\\x=1.65 \ and\ x=1.35

The reaction time between 1.35 and 1.65 seconds

8 0
3 years ago
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zheka24 [161]

Answer:

1

Explanation:

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3. 1

Fe. (OH)

Fe(OH)3

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3 years ago
What type of energy is a furnace and using a telephone? pls help
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Answer:

hope this helped

Explanation:

Radiant energy is created through electromagnetic waves and was discovered in 1885 by Sir William Crookes. Fields in which this terminology is most often used are telecommunications, heating, radiometry, lighting, and in terms of energy created from the sun.

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3 years ago
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A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w
Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)

c=5760J/^oC

\Delta T=0.570^oC

To calculate the enthalpy change, we use the formula:

\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = \frac{3283.2J}{0.1326g}=24760.181J/g

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol

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Oxidation is _______?
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