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kkurt [141]
3 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]3 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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alukav5142 [94]

Answer:

A.

Explanation:

Adjust = Adapt if that makes sense.

8 0
2 years ago
Read 2 more answers
Brainliest for an answer!
Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0
2 years ago
The reaction of hydrogen gas and iron oxide is described by the chemical equation below. 3H2+Fe2O3→2Fe+3H2O How many moles of ir
ELEN [110]

Answer:

2.2 moles of Fe will be produced

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 3.3 moles

Number of moles of iron oxide = 1.5 moles

Step 2: The balanced equation

3H2 + Fe2O3 → 2Fe + 3H2O

Step 3: Calculate the limiting reactant

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles

There will remain 1.5 - 1.1 = 0.4 moles Fe2O3

Step 4: Calculate moles Fe

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe

2.2 moles of Fe will be produced

5 0
3 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n
ArbitrLikvidat [17]

Answer:

\boxed{\text{3.3 mL}}

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}

2. Moles of NaOH  

\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}

3. Molar concentration of NaOH

c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

c_{1}V_{1} = c_{2}V_{2}

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1   mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}

(b) Calculate the volume  of dilute solution

\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}

\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}

4 0
3 years ago
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