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Nataliya [291]
3 years ago
5

Solution A and solution B are separated by a semipermeable membrane. Solution A contains 1% glucose, solution B contains 5% gluc

ose. By diffusion: Solution A and solution B are separated by a semipermeable membrane. Solution A contains 1% glucose, solution B contains 5% glucose. By diffusion: water will move from solution A to solution B. glucose will move from solution B to solution A. water will move from solution B to solution A. glucose will move from solution A to solution B.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

Glucose will move from the solution B to the solution A

Explanation:

Given that:

Solution A contains 1% glucose, and,

Solution B contains 5% glucose

Diffusion is the net movement of the substance from the region of the higher concentration to the region of the lower concentration.

Thus, solution B contains more concentration of glucose as compared to solution A. <u>By the process of diffusion, the particle moves from higher concentration to lower concentration and thus, glucose will move from solution B to solution A.</u>

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A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a tempera
Hitman42 [59]

Answer:

The final volume is 3.07L

Explanation:

The general gas law will be used:

P1V1 /T1 = P2V2 /T2

V2 =P1 V1 T2 / P2 T1

Give the variables to the standard unit:

P1 = 345 torr = 345 /760 atm = 0.4539atm

T1 = -15°C = -15 + 273 = 258K

V1 = 3.48L

T2 = 36°C = 36+ 273 = 309K

P2 = 468 torr = 468 * 1/ 760 atm = 0.6158atm

V2 = ?

Equate the values into the gas equation, you have:

V2 = 0.4539 * 3.48 * 309 / 0.6158 * 258

V2 = 488.0877 /158.8764

V2 = 3.07

The final volume is 3.07L

8 0
3 years ago
A sample of gas has a volume of 100.0 L at 135°C. Assuming the pressure remains constant, what is the volume of the gas if its t
lina2011 [118]

Answer: 84.56L

Explanation:

Initial volume of gas V1 = 100L

Initial temperature T1 = 135°C

Convert temperature in Celsius to Kelvin

( 135°C + 273 = 408K)

Final temperature T2 = 72°C

( 72°C + 273= 345K)

Final volume V2 = ?

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

100L/408K = V2/345K

To get the value of V2, cross multiply

100L x 345K = V2 x 408K

34500 = V2 x 408K

V2.= 34500/408

V2 = 84.56L

Thus, the volume of the gas becomes 84.56 liters

8 0
3 years ago
How much volume (in cm^3)is gained by a person who gains 11.8 lb of our fat? Human fat has a density of 0.918 g/cm^3
Zinaida [17]

The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert pounds to grams </h3>

1 lb = 453.592 g

Therefore,

11.8 lb = 11.8 × 453.592

11.8 lb = 5352.3856 g

<h3>How to determine the volume </h3>
  • Mass = 5352.3856 g
  • Density = 0.918 g/cm³
  • Volume =?

Volume =  mass / density

Volume =  5352.3856 / 0.918

Volume = 5830.49 cm³

Learn more about density:

brainly.com/question/952755

#SPJ1

3 0
2 years ago
Which of these solutions are basic at 25 °C? Solution A: [OH−]=3.13×10−7 M Solution C: [H3O+]=0.000747 M Solution B: [H3O+
solong [7]

Answer:

Are basic:

[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M

Explanation:

A solution is basic when pH = - log [H₃O⁺] is higher than 7.

It is possible to convert [OH⁻] to [H₃O⁺] using:

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

a. [OH⁻] = 3.13x10⁻⁷M

[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]

[H₃O⁺] = 3.19x10⁻⁸M

pH = - log [H₃O⁺] = 7.50

[OH⁻] = 3.13x10⁻⁷M is basic

b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.

This solution is not basic

c. [H₃O⁺] = 9.55x10⁻⁹M

pH = 8.02

This solution is also basic.

8 0
2 years ago
I need help ASAP 10 points
jekas [21]

Answer:

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