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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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Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

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4 years ago
When writing an electron configuration for a transition metal, what is the last orbital type to be filled – an s, p, d or f orbi
Black_prince [1.1K]
It would be d.

Reason being said...

the electron configuration normally goes like this...

1s2 2s2 2p6 3s2 3p6 4s2....

until you hit the transition metals..remember those have a special rule..

even though you are in the 4 sublevels for the orbitals ... it goes down 1

Making it 3d..(1,2,3,4,5,6,7,8,9,10)

Going on...

at 5s2 then, 4d1, 4d2, 4d3, 4d4, etc..

at 6s2 then, 5d1, 6d2, 6d3, 6d4, etc..

Thus, D orbital is your answer.
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Answer:

Dióxido de azufre

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