How many atoms of ^6Li are there in 12.3 g of ^6Li?

If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln

Ksju [112]

Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.

5 0

2 years ago

Read 2 more answers

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

A sample of pure iron has a mass of 5.00 g. Calculate its volume.

olga55 [171]

<span>It's volume is 0.48 cm3Specific </span>

3 0

2 years ago

In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used

DENIUS [597]

Answer:

A. Water and Sugar can be separated by evaporation and then crystallization

B. Mixture of Hexane and Octane can be separated by distillation

C. Solid Iodine, I₂ and NaCl can be separated by filtration and then evaporation

D. "Sharpie" permanent marking pen can be separated by chromatography

E. Nickel shavings and copper pellets can be separated by magnetic separation

Explanation:

A. A mixture of water and sugar can be separated by employing two separation techniques, evaporation and crystallization. First the sugar solution is heated to evaporate most of the water. When the solution becomes very saturated, it is allowed to cool and then the sugar molecules are obtained through crystallization induced by seeding or scratching the walls of the container.

B. A mixture of hexane (boiling point = 68 °C) and Octane (boiling point = 125 °C) can be separated by distillation due to their significant difference in boiling points.

The mixture is heated in a flask connected to a Liebig condenser. Hexane with the lower boiling point will distill over first and is collected. Afterwards, octane next distills over and is collected as well.

C. A mixture of solid iodine and NaCl can be seperated by first dissolving in water. Iodine being non- polar does not dissolve and is collected as a residue from filtration using a filter paper, while the NaCl solution is collected as the filtrate. The NaCl is recovered from solution by evaporating to dryness in an evaporating dish.

D. "Sharpie" permanent marking pen contains a mixture of dyes which can be separated by paper chromatography.

A drop of the marker ink is placed on a spot above the solvent level on the paper strip used for the separation. The paper strip is held vertically inside a jar containing a solvent which serves as the mobile phase. The jar is covered and the different dyes move along the paper which serves as the stationary phase, and is thus separated. The paper strip is removed from the jar when the ascending front of the solvent is approaching the top of the paper. The paper is dried and the various dyes can be identified by comparing the distance each has traveled with those of standards.

E. A mixture of nickel shavings and copper pellets can be separated by magnetic separation.

A magnet is brought near the mixture and the nickel shavings being magnetic is attracted to the magnet leaving copper pellets behind since copper is not magnetic.

4 0

2 years ago

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of

swat32

Ideal solutions obey Raoult's law, which states that:

P_i = x_i*(P_pure)_i

where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i

In this case,

P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr

The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:

P_total = (439.6 + 118.9) torr = 558.5 torr

Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:

118.9 torr/558.5 torr = 0.213

8 0

2 years ago