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alexira [117]
3 years ago
11

Ascorbic acid, or vitamin C (C6H8O6, molar mass = 176 g/mol), is a naturally occurring organic compound with antioxidant propert

ies. A healthy adult’s daily requirement of vitamin C is 70-90 mg. A sweet lime contains 2.82×10−4 mol of ascorbic acid.
To determine whether the ascorbic acid in a sweet lime meets the daily requirement, calculate the mass of ascorbic acid in 2.82×10−4 mol of ascorbic acid.
Chemistry
2 answers:
Rus_ich [418]3 years ago
6 0

Answer:

The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.

Explanation:

To determine the mass of ascorbic acid knowing the number of moles we use the following formula:

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of ascorbic acid = 2.82 × 10⁻⁴ × 176

mass of ascorbic acid = 496 × 10⁻⁴ g = 0.0496 g = 49.6 mg

daily requirement of ascorbic acid = 70 - 90 mg

The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.

Bumek [7]3 years ago
4 0

<u>Answer:</u>

<em>49.7 mg of Ascorbic acid do not meet the daily requirement.</em>

<em></em>

<u>Explanation:</u>

The conversions are  

Moles to mass, by multiplying with molar mass.

Molar mass is the mass of 1 mole of the substance.

Its unit is gram per mole g/mol

Ascorbic acid (C_6H_8O_6) contains 6, C atoms 8, H atoms and 6, O atoms .

We find the molar mass of the compound by just adding the atomic mass of the atoms present in it

Molar mass of Ascorbic acid

=(6\times 12.0107)+(8\times 1.00784)+(6\times 15.999)

=176.124 g/mol(Given)

Mass of Ascorbic acid = Moles × Molar mass

=2.82 \times 10^{-4} mol \times 176.124 g/mol

=0.0497 g

We know

1 g = 1000 mg

So

0.0497g \times 1000mg/1g=49.7mg VITAMIN C

The required quantity for an Adult is 70 to 90mg.

<em>So 49.7 mg of Ascorbic acid do not meet the daily requirement.(Answer)</em>

<em></em>

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Explanation:

B. Ester is R-CO-OR

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When oxygen is available what happens immediately after glycolysis?
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3 0
3 years ago
In naming the compound PCl5, the prefix used with the second element is ____________________.
Oksanka [162]

Answer:

\boxed {\boxed {\sf Hepta}}

Explanation:

We are given the formula:

PCl_5

This is a molecular formula, because it contains nonmetals.

1. Name the first element

The first element is phosphorous (P). Since this is the first element and there is only one, we don't need a prefix.

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2. Second element

The second element is chlorine (Cl). It has a subscript of 5, so we must add the prefix of <u>hepta</u>-.

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6 0
3 years ago
A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO
labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

5 0
3 years ago
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Answer:

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