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Arada [10]
2 years ago
3

You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/

mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and
Chemistry
1 answer:
Pavlova-9 [17]2 years ago
4 0

Answer : The number of drops pf the stock solution mist be added are, 50 drops.

Explanation :

As we are given that the concentration of stock solution 1 mg/mL and pipette delivers 20 drops/mL. That means,

1 mg of epinephrine = 1 mL = 20 drops

The final volume of Locke's solution = 25 mL

and the final concentration needed = 100\mu g/mL

As, 1 mL of solution contains = 100\mu g

So, 25 mL of solution contains = \frac{25mL}{1mL}\times 100\mu g=2500\mu g=2.5mg

Conversion used : 1\mu g=0.001mg

Now we have to determine the number of drops needed.

As, 1 mg of epinephrine contains 20 drops.

So, 2.5 mg of epinephrine contains 2.5 × 20 = 50 drops.

Therefore, the number of drops pf the stock solution mist be added are, 50 drops.

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d.  withdraws electrons inductively

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c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

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<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=9.06g

Mass of H_2O=5.58g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4

Thus, the empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

3 0
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