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Sergeeva-Olga [200]
3 years ago
12

A standard solution is prepared for the analysis of fluoxymesterone (c20h29fo3; 336 g/mol), an anabolic steroid. a stock solutio

n is first prepared by dissolving 16.8 mg of fluoxymesterone in enough water to give a total volume of 500.0 ml. a 1.0 μl aliquot (portion) of this stock solution is taken out and then diluted to a final volume of 1.0 ml. calculate the final concentration of this new solution in m.
Chemistry
1 answer:
-Dominant- [34]3 years ago
3 0

<em>Answer:</em>

  • The concentration of new solution will be 1×10∧-7 M.

<em>Solution:</em>

<em>Data Given </em>

       given mass of fluoxymesterone =16.8mg = 0.0168 g

       molar mass  of fluoxymesterone = 336g/mol

       vol. of fluoxymesterone = 500.0 ml = 0.500 L

      Stock Molarity of  fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M

So applying dilution formula

                   Stock Solution :  New Solution

                                 M1.V1 = M2.V2

       ( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L

     [( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2

     1 × 10∧-7 = M2

<em>Result:</em>

  • The concentration of new solution M2 will be  1 × 10∧-7
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Glucose, C6H12O6,C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the eq
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Answer:

Mass of oxygen = 61.824 g

Mass of carbon dioxide = 85.01 g

Explanation:

Given data:

Mass of glucose = 58 g

Mass of carbon dioxide = ?

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Solution:

First of all we will write the balanced chemical equation,

C₆H₁₂O₆  + 6O₂       →    6CO₂  + 6H₂O

Moles of glucose:

Number of moles = mass / molar mass

Number of moles = 58 g/180 g/mol

Number of moles =  0.322 mol

Now we compare the moles of oxygen with glucose from balance chemical equation.

                             C₆H₁₂O₆          :              O₂  

                                    1                :              6

                                    0.322       :              0.322×6 = 1.932 mol

Mass of oxygen:

Mass of oxygen = number of moles × molar mass

Mass of oxygen =  1.932 mol × 32 g/mol

Mass of oxygen =  61.824 g

Now we compare the moles of carbon dioxide with moles of glucose and oxygen.

                              C₆H₁₂O₆            :              CO₂

                                   1                   :                 6

                                  0.322           :           0.322×6 = 1.932 mol

                                   

                                 O₂                    :                 CO₂

                                  6                     :                  6

                                 1.932                :                  1.932

Mass of carbon dioxide;

mass of carbon dioxide = number of moles × molar mass

mass of carbon dioxide =  1.932 mol × 44 g/mol

mass of carbon dioxide =  85.01 g

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You have the reactants who react to make your products.

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The limiting reactant is the reactant that will RUN OUT FIRST and that will establish the maximum amount of product that will be produced.

To make an example, let's look at this equation and at the following question:

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We know that 2Al + 6HCL are the reactants, right?

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But, how do we know which one is the limiting reactant?

To know which one is the limiting reactant we need to know HOW MUCH H2 can Al produce, and also how much H2 can HCL produce. Why?

It's simple, because if you find the one that produces the less amount of H2, you know immediately that's the maximum amount of that product that will be produced.

Let's say you have a sandwich that needs to be made with 2 slices of bread, 3 meats and 1 cheese.

But you have got 4 slices of bread, 9 slices of meat, and 5 slices of cheese.

Well, you could make 2 sandwiches with 4 slices of bread, 3 sandwiches with those 9 slices of meat and 5 sandwiches with the 5 slices of cheese.

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Coming back to our equation, we can find the number of moles of H2 produced by each one of the reactants, Al and HCl. I'll find the number of moles quickly to show you what the concept of limiting reactant is.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As you can see, 6.5 mol of H2 is the maximum that can be produced by the HCL and is less than the Al, so that maximum amount that you will get in the product H2 is no more than 6.5.

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3 years ago
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