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3 years ago

Prepare 15 mg/dl working standard solution from a stock solution of 20 mg/dl. State the volume of diluent and dilution.

1 answer:

4 0

Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.

Recalling the principle:

initial volume x initial molarity = final volume x final molarity

Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:

Applying the dilution principle:

10 x 20 = final volume x 15

final volume = 200/15

= 13.33 mL

This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.

More on dilution principle can be found here: brainly.com/question/11493179

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Through what type of single pericyclic reaction did the reaction shown below proceed

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A. [1,2] sigmatropic hydrogen migration

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3 years ago

What does weight measure?

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4 years ago

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Name physical properties

Paraphin [41]

Answer:

Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils.

melting point (intensive): the temperature at which a substance melts.

Explanation:

Hope this helped! <3

8 0

3 years ago

The balloon in the previous problem will burst if its volume reaches 400. L. Given the initial conditions specified in that prob

NARA [144]

This is an incomplete question, here is complete question.

A metrological balloon contains 250 L of He at 22 C and 740 mmHg.

The balloon in the previous problem will burst if its volume reaches 400 L. Given the initial conditions specified in that problem, at what temperature, in degrees Celsius, will the balloon burst if its pressure at that bursting point is 0.475 atm.

Answer : The final temperature will be, $-44.4^oC$

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 740 mmHg = 0.974 atm

$P_2$ = final pressure of gas = 0.475 atm

$V_1$ = initial volume of gas = 250 L

$V_2$ = final volume of gas = 400 L

$T_1$ = initial temperature of gas = $22^oC=273+20=293K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{0.974atm\times 250L}{293K}=\frac{0.475atm\times 400L}{T_2}$

$T_2=228.6K=228.6-273=-44.4^oC$

Thus, the final temperature will be, $-44.4^oC$

7 0

4 years ago

Percent of water in Na2CO3 • 10H2O

Troyanec [42]

Hey there!

In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage.

→ Na₂CO₃ × 10 H₂O

→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15

Now, just add all of those numbers up for the total molar mass.

→ 45.9796 + 12.0107 + 47.997 + 180.15 = 286.1373

The last step is to divide the molar mass of the 10 water molecules by the total mass.

→ 180.15 ÷ 286.1373 = 0.62959 ≈ 0.63

Your answer will be about 63%.

Hope this helped you out! :-)

5 0

3 years ago