According to Dalton, atoms cannot be SELECT ALL THAT APPLY a Created b Destroyed c Subdivided d Bonded

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

PSYCHO15rus [73]

This question is incomplete, the complete question is;

A certain liquid X has a normal boiling point of 150.4 °C and a molar boiling point elevation constant kb is 0.60 °Ckgmol⁻¹.

A solution is prepared by dissolving some urea (NH22CO) in 750 g of X. This solution boils at 150.9 °C . Calculate the mass of urea that was dissolved. Round your answer to 3 significant digits.

Answer:

the mass of urea that was dissolved is 37.5 g

Explanation:

Given the data in the question;

normal boiling point of X; Tb⁰ = 150.4 °C

boiling point of solution Tb = 150.9 °C

Change in boiling point Δt = Tb - Tb⁰ = 150.9 °C - 150.4 °C = 0.5 °C

Kb = 0.6 °C.kg.mol⁻¹

V = 750 g

Now, we know that

Δt = Kb × molality

so

0.5 = 0.6 × molality

molality = 0.5 / 0.6

molality = 0.833

we know that molar mass of urea is 60 g/mol

so

molality = mass × 1000 / molar mass × volume( g )

we substitute

0.833 = ( mass × 1000 ) / ( 60 × 750 )

0.833 = ( mass × 1000 ) / 45000

0.833 × 45000 = mass × 1000

mass = ( 0.833 × 45000 ) / 1000

mass = 37485 / 1000

mass = 37.485 ≈ 37.5 g { 3 significance figure }

Therefore, the mass of urea that was dissolved is 37.5 g

8 0

2 years ago

Proteins are large molecules. They enter a cell through

Lady bird [3.3K]

The answer is; active transport

This means that energy is expended in the process. Because large protein cannot be transported through protein channels in the plasma membrane for being too big, they are transported through endocytosis. This means a vesicle merges with the plasma membrane and forms an invagination that engulfs the protein. The vesicle then encloses with the protein inside and moves into the cell.

7 0

3 years ago

Enter your answer in the provided box. an aqueous solution containing 10 g of an optically pure substance was diluted to 500 ml

sladkih [1.3K]

Answer:

[α] = -77.5° / $\frac{\textup{dm-g}}{\textup{mL}}$

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

Volume of water = 500 mL

Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm

measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = $\frac{\alpha}{c\times l}$

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = $\frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}$

or

c = $\frac{10}{500}$

or

c = 0.02 g/mL

on substituting the values, we get

[α] = $\frac{-3.10^o}{0.02\times2}$

or

[α] = -77.5° / $\frac{\textup{dm-g}}{\textup{mL}}$

7 0

3 years ago

PLEASE HELP<br> I really need someone to answer this!!!!

marshall27 [118]

Answer: I think it’s the first one

6 0

3 years ago

Read 2 more answers

Two identical 732.0 L tanks each contain 212.0 g of gas at 293 K, with neon in one tank and nitrogen in the other. Based on the

Vladimir [108]

Answer: (a) Neon, Nitrogen; (b) Neon, Nitrogen; (c) Neon is lower than Nitrogen; (d) It doesn't affect;

Explanation: The kinetic-molecular theory studies the behavior of particles under pre-determinated situation. In cases of gases, the particles moving around colliding with each other and the walls of the container, without loss of energy. In the case in question, all the parameters are the same (same temperature, volume and pressure), except for the gases, which has different molar masses. In this sense, Neon has lower average speed due to its molar mass being higher, which means, its particles moves slower for being heavier. Related to pressure, as velocity is lower, it collides less with the walls of the tank, and so pressure is lower. For density, it doesn't affect the behavior of the system nor the kinetic energy.

7 0

3 years ago

According to Dalton, atoms cannot be SELECT ALL THAT APPLY a Created b Destroyed c Subdivided d Bonded​

According to Dalton, atoms cannot be SELECT ALL THAT APPLY a Created b Destroyed c Subdivided d Bonded