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3 years ago

How many mL of a stock solution of 2.00 M KNO3 are needed to prepare 100.0 mL of 0.15M KNO3? with work plz

1 answer:

3 0

Using the law of dilution :

Mi x Vi = Mf x Vf

2.00 x Vi = 0.15 x 100.0

2.00 x Vi = 15

Vi = 15 / 2.00

Vi = 7.5 mL

hope this helps!

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A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

3 years ago

The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o

IRISSAK [1]

Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

b = mass of solute / molar mass of solute × kilograms of solvent

b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg

Step 3: Calculate the molal freezing point depression constant Kf of X

Freezing point depression is a colligative property. It can be calculated using the following expression.

ΔT = Kf × b

Kf = ΔT / b

Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol

7 0

3 years ago

Is syrup a compound or solution

S_A_V [24]

Answer:

Syrup is a mixture of thick sticky liquid consisting of a concentrated solution of sugar and water with or without the addition of a flavoring agent or medicinal substance.

Explanation:

so compound

5 0

3 years ago

Which could be the pH of an aqueous solution with a H3O+ ion concentration that is less than its OH- ion concentration?

anzhelika [568]

Answer:

Explanation:

You can think of pH as "parts Hydrogen ion," but remember that the pH scale is "backwards." The pH scale ranges from 0 to 14, with zero being the most acidic (highest concentration of H+) and 14 being the most basic.

4 0

2 years ago

Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi

JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>

Mass of CO₂ = 1.47 g
Molar mass of CO₂ = 44 g/mol
Molar of C = 12 g/mol
Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>

Mass of compound = 0.5 g
Mass of C = 0.4 g
Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>

C = 0.4 g
H = 0.1 g
Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>

Empirical formula = CH₃
Molar mass = 30 g/mol
Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

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4 0

2 years ago