Ascorbic acid, or vitamin c (c6h8o6, molar mass = 176 g/mol), is a naturally occurring organic compound with antioxidant properties. 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.

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.

49.7 mg of Ascorbic acid do not meet the daily requirement.

Explanation:

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 () 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

=176.124 g/mol(Given)

Mass of Ascorbic acid = Moles × Molar mass

=0.0497 g

We know

1 g = 1000 mg

So

VITAMIN C

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

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

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o