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2 years ago

In a solution with a pH of 3.0, the color of

Chemistry

1 answer:

Alinara [238K]2 years ago

5 0

Answer:

A) litmus is red

Explanation:

To answer this question, it can be helpful to have the color charts. Litmus, phenolphthalein and methyl orange are ways to test the pH of a substance.

Litmus paper

Litmus can tell you if a substance is an acid or a base. You need to put the substance on both red litmus and blue litmus paper.

pH < 7: both papers are red. 3.0 is less than 7.

pH = 7: none of them change color

pH > 7: both papers are blue

Phenolphthalein

When this indicator is added to a substance, the result is either colorless or pink.

0 < pH ≤ 7: colorless. The color is not red or blue for pH 3.0.

pH > 7: pink

Methyl orange

0 < pH < 4: red. The color is not yellow if the pH is 3.0.

4 ≤ pH < 5: orange

pH ≥ 5: yellow

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Review the equation below. 2KClO3 mc012-1.jpg 2KCl + 3O2 How many moles of oxygen are produced when 2 mol of potassium chlorate

Vlada [557]

Stoichiometry of the reaction:

2 KClO₃ = 2 KCl + 3 O₂
 ↓ ↓

2 mole KClO₃ ----------> 3 mole O₂
2 mole KClO₃ ----------> ?

KClO₃ = 2 * 3 / 2

KClO₃ = 6 / 2

= 3 moles de KClO₃

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3 years ago

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2 years ago

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2 years ago

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

3 0

3 years ago