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3 years ago

If all of your calculations involved using a 'hot pink' result, what would be the result of the unknown acid concentration?

Chemistry

1 answer:

Ket [755]3 years ago

7 0

Answer:

it would appear to be more concentrated than it should be because more base was added that should have been

Explanation:

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In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 gram

kherson [118]

This question is incomplete, the complete question is;

In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization

Compound Solubility in water at 25°C Solubility in water at 100°C

Benzoic Acid 0.34 g/100mL 5.6 g/100mL

Acetanilide 0.53 g/100mL 5.5 g/100mL

Answer:

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

Explanation:

Given the data in the question;

mass of sample = 5 g

percentage composition of the acetanilide impurity = 6.3%

mass of the acetanilide in impure sample will be;

⇒ 6.3% × 5 g = 0.315 g

Mass of benzoic acid in impure sample;

⇒ 5 g - 0.315 g = 4.685 g

now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL

hence 4.685 g of benzoic acid is soluble in x mL

x = [ 100 mL × 4.685 g ] / 5.6 g

x = 83.66 ≈ 84 mL

Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL

hence 0.315 g of benzoic acid is soluble in x mL

x = [ 100 mL × 0.315 g ] / 5.5 g

x = 5.727 ≈ 6 mL

So, the minimum amount in mL of solvent (water) required for the recrystallization will be;

⇒ 85 mL + 6 mL = 91 mL

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

5 0

3 years ago

What is represented at the triple point of water?

Aleks04 [339]

Answer:D

Explanation: Solid, liquid, and gas phases in equilibrium

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$