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Novosadov [1.4K]
2 years ago
10

If you want to produce 75,000 bottles of 200 mg vitamin C pills, and there are 100 pills per bottle, how many kg of vitamin C do

you need?
Chemistry
1 answer:
Rasek [7]2 years ago
3 0

Answer:

200–300 mg/day ascorbic acid from vitamin C-rich foods [10]. ... as high as 3 g ascorbic acid taken every 4 hours would produce peak ... individuals should ensure that they meet the RDA for vitamin C [4,8].

Explanation:

You might be interested in
Equation Unbalanced : __AlCl3 + __F2 --> __AlF3 + __Cl2
Greeley [361]

\\ \rm\Rrightarrow 2AlCl_3+3F_2\longrightarrow 2AlF_3+3Cl_2

  • Balanced

3mols of fluorine produce 2 mol chlorine

1 mol fluorine produces 2/3=0.6mol chlorine

Moles of fluorine gas=

  • 4/19
  • 0.2mol.approx

Moles of chlorine:-

  • 0.2(0.6)=0.18mol

Mass of chlorine

  • 71(0.18)
  • 12.78g
6 0
2 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
8 Fe + Sg → 8 Fes<br><br> How many grams of FeS is produced from 0.3 mol Sg?
frozen [14]

Answer:

Explanation:

Molar ratio for Sg : FeS = 1:8

If there are 0.3 moles for Sg

Therefore, 0.3 × 8 =2.4 moles of FeS

Mass = Moles/ Mr

Mr of FeS = 56+32=88

So mass = 2.4/88

Mass= 0.027g

6 0
2 years ago
A piece of copper has a mass of 660 g. How many atoms does the sample contain?
son4ous [18]
The conversions that will be used are:

1 mole of copper / 63.5 grams of copper
6.02 x 10²³ atoms of copper / 1 mole of copper

Multiplying the given mass by these conversions,

660 g * (1 mol Cu / 63.5 g Cu) * (6.02 x 10²³ Cu atoms / 1 mole Cu)

The sample contains 6.25 x 10²⁴ atoms of copper
4 0
3 years ago
which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4
olasank [31]

Answer:

Answer: CH₃ and C₂H₆ have same empirical formula.

Explanation:

it just compares in that it its the same

7 0
3 years ago
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