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kkurt [141]
4 years ago
10

A student has 100. mL of 0.400 M CuSO4(aq) and is asked to make 100. mL of 0.150 M CuSO4(aq) for a spectrophotometry experiment.

The following laboratory equipment is available for preparing the solution: centigram balance, weighing paper, funnel, 10 mL beaker, 150 mL beaker, 50 mL graduated cylinder, 100 mL volumetric flask, 50 mL buret, and distilled water.
a. Calculate the volume of 0.400 M CuSO4(aq) required for the preparation
b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO4(aq) from the 0.400 M CuSO4(aq)
Chemistry
1 answer:
ValentinkaMS [17]4 years ago
3 0

Answer:

<u></u>

  • <u>a. 37.5 mL</u>

<u></u>

  • <u>b. See the description below</u>

Explanation:

<u><em>a. Volume of 0.400 M CuSO₄(aq) required for the preparation</em></u>

In dissolutions, since the number of moles of solute is constant, the equation is:

          C_1\times V_1=C_2\times V_2

Substitute and solve for V₁

         0.400M\times V_1=0.150M\times 100ml

          V_1=0.150M\times 100.ml/0.400M=37.5ml

<u><em>b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO₄(aq) from the 0.400 M CuSO₄(aq)</em></u>

You will use the stock solution, the funnel, the buret, and distilled water.

i) Using the funnel, fill in the buret with with 50 ml of the stock solution, i.e. the 100. ml of 0.400 M CuSO₄(aq) solution.

ii) Pour 37.5 ml of the stock solution from the burete into the volumetric flask.

iii) Carefully add disitlled water to the 37.5ml of the stock solution in the volumetric flask until the mark (50 ml) on the volumetric flask.

iv) Put the stopper and rotate the volumetric flask to homegenize the solution.

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Plants undergo photosynthesis to produce glucose according to the reaction below. What mass of water is required to produce 5.0g
solniwko [45]

Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

                   6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

                        6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

                                     6 mol               1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆  = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆  = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

             6CO₂      +  6H₂O          --------->     C₆H₁₂O₆   +    6O₂

                              6 mol (18 g/mol)           1 mol (180 g/mol)

                                  108 g                            180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

So

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

               108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

                X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

                     mass of water (H₂O) = 108 g x 5 g / 180 g

                     mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.  

7 0
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A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your
amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically,         Molarity = \frac{\text{no. of moles}}{\text{Volume in liter}}

Also, when number of moles are equal in a solution then the formula will be as follows.

                     M_{1} \times V_{1} = M_{2} \times V_{2}

It is given that M_{1} is 8.00 M, V_{1} is 7.00 mL, and M_{2} is 0.80 M.

Hence, calculate the value of V_{2} using above formula as follows.

                    M_{1} \times V_{1} = M_{2} \times V_{2}

                 8.00 M \times 7.00 mL = 0.80 M \times V_{2}

                      V_{2} = \frac{56 M. mL}{0.80 M}

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