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3 years ago

What is the molarity of a 3.0-liter solution that contains 0.45 moles of solute

2 answers:

6 0

Moles = n/v where n is the moles of solute and v being the liters of solution.
We can put in the information provided to find the molarity.

Moles = .45/3.0 = .15
So we now know that the molarity of that solution is .15!
I hope I helped you :). Make sure to memorize that formula because it's not that hard as long as you know what to plug in.

Ksivusya [100]3 years ago

5 0

Answer : The molarity of the solution is, 0.15 mole/L

Solution :

Molarity : It is defined as the number of moles of solute present in one liter of the solution.

Formula used :

$Molarity=\frac{\text{moles of solute}}{\text{Volume of solution in liters}}$

Now put all the given values in this formula, we get the molarity of the solution.

$Molarity=\frac{0.45mole}{3L}=0.15mole/L$

Therefore, the molarity of the solution is, 0.15 mole/L

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What does Le Châtelier's principle state?

bonufazy [111]

When a system experiences a disturbance ( such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

5 0

3 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

Answer: The freezing point of solution is -0.974°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

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

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp

maksim [4K]

Answer:

Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: Volume of CO₂ produced = 1223.21 L

Note: the complete second part of the question is given below:

2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Explanation:

Part 1: Balanced chemical equation

C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

From the equation of the reaction;

At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L

Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.

O.470 Kg of nonane = 470 g of nonane

470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

3 0

3 years ago

In each pair, choose the larger of the indicated quantities or state that the samples are equal:

romanna [79]

There are more total ions in 2.2 moles of $MgCl_2$

Ion, any atom or group of atoms that bears one or more positive or negative electrical charges. Positively charged ions are called cations; negatively charged ions, anions.

Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the formerly bonded atoms.

Examples of these processes include the reaction of a sodium atom with a chlorine atom to form a sodium cation and a chloride anion; the addition of a hydrogen cation to an ammonia molecule to form an ammonium cation; and the dissociation of a water molecule to form a hydrogen cation and a hydroxide anion.

Learn more about ions

brainly.com/question/13692734

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8 0

1 year ago

If 38.5 grams of potassium react with excess oxygen gas, how many grams of potassium oxide can be produced? 4K + O2 yields 2K2O

Lera25 [3.4K]

Answer:

46.40 g.

Explanation:

It is a stichiometric problem.

The balanced equation of the reaction: 4K + O₂ → 2K₂O.

It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.

We should convert the mass of K (38.5 g) into moles using the relation:

n = mass / molar mass,

n = (38.5 g) / (39.098 g/mol) = 0.985 mole.

Using cross multiplication:

4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.

0.985 mole of K produces → ??? moles of K₂O.

∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.

Now, we can get the mass of K₂O:

∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.

6 0

3 years ago