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solniwko [45]
2 years ago
10

what is the new concentration of A if 7.7 ml of a .10 M solution of A is mixed with 18.3 mL of .20 M solution B and an additiona

l 11.7 ml water?
Chemistry
1 answer:
Elenna [48]2 years ago
4 0

Answer:

0.12 M

Explanation:

To find the new concentration, you need to (1) find the moles of Solution A (via the molarity equation), then (2) find the moles of Solution B, then (3) combine all of the moles and volumes, and then (4) calculate the new molarity. The final answer should have 2 sig figs to reflect the lowest amount of sig figs in the given values.

<u>Solution A:</u>

7.7 mL / 1,000 = 0.0077 L

Molarity = moles / volume

0.10 M = moles / 0.0077 L

0.00077 = moles

<u>Solution B:</u>

18.3 mL / 1,000 = 0.0183 L

Molarity = moles / volume

0.20 M = moles / 0.0183 L

0.00366 = moles

<u>New Solution:</u>

11.7 mL / 1,000 = 0.0117 L

New Volume = 0.0077 L + 0.0183 L + 0.0117 L

New Volume = 0.0377 L

New Moles = 0.00077 moles + 0.00366 moles

New Moles = 0.00443 moles

Molarity = moles / volume

Molarity = 0.0043 moles / 0.0377 L

Molarity = 0.1175 M = 0.12 M

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The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl
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Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of  caffeine extracted is  P =  39.8 \ mg

Explanation:

From the question above  we are told that

    The  K value is  K =  4.6

     The  mass of the caffeine is  m  = 40 mg

      The  volume of water is  V  = 2 mL

      The volume  of caffeine is  v_c =  2 mL

     The number of times the extraction was done is  n =  3

Generally the mass of  caffeine that will be  extracted is  

           P =  m  *  [\frac{V}{K *  v_c + V} ]^3

substituting values  

           P =  40   *  [\frac{2}{4.6 *  2 + 2} ]^3

           P =  39.8 \ mg

6 0
3 years ago
If a radioactive material has a 10 year half-life, how much of a 100 g sample will be left after 30 years?
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12.5g, each 10 years you lose a half of what you have at that given moment

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2 years ago
8Which of the following can be found in the Kuiper Belt?
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Explanation:

8 0
3 years ago
Read 2 more answers
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
2 years ago
A vessel contains a mixture of gases. The mass of each gas used to make the mixture is known. Which of the following information
faltersainse [42]

Answer: The molar mass of each gas

Explanation:

Mole fraction is the ratio of moles of that component to the total moles of solution. Moles of solute is the ratio of given mass to the molar mass.

\text{Mole fraction of solute}=\frac{\text{Moles of solute}}{\text{Total Moles}}

Suppose if there are three gases A, B and C.

a) \text{Mole fraction of A}=\frac{\text{Moles of A}}{\text{ Moles of (A+B+C)}}

b) \text{Mole fraction of B}=\frac{\text{Moles of B}}{\text{ Moles of (A+B+C)}}

c) \text{Mole fraction of C}=\frac{\text{Moles of C}}{\text{ Moles of (A+B+C)}}

moles of solute =\frac{\text {given mass}}{\text {Molar mass}}

Thus if mass of each gas is known , we must know the molar mass of each gas to know the moles of each gas.

8 0
3 years ago
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