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3 years ago

11

The difference between simple diffusion and facilitated diffusion is that facilitated diffusion

1 answer:

Sholpan [36]3 years ago

6 0

Answer:

Simple and facilitated diffusion are both similar types of phenomenon where the substances are moved from a region of high concentration to lower concentration across a semi-permeable membrane.

The main difference between them is that the facilitated diffusion requires a special type of protein or carrier molecules in order to transport the substances whereas simple diffusion does not require such molecules. The rate of flow of these substances is high in the facilitated diffusion in comparison to the simple diffusion.

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Which atoms make up a molecule of sodium bicarbonate.

Ratling [72]

Answer:

Its chemical formula is NaHCO3. Its formula consists of one sodium (Na) atom, one hydrogen (H) atom, one carbon (C) atom and three oxygen (O) atoms.

Explanation:

Hope this helps

8 0

2 years ago

A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity?

Softa [21]

Answer:

1,100 km/h

Explanation:

Velocity = distance/time = 4,400 km / 4.0 h = 1,100 km/h

3 0

4 years ago

Read 2 more answers

A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant Kf = 5.32 °C.

Citrus2011 [14]

Answer : The freezing point of a solution is $-15.4^oC$

Explanation : Given,

Molal-freezing-point-depression constant $(K_f)$ = $5.32^oC/m$

Mass of urea (solute) = 29.82 g

Mass of solvent = 500 g = 0.500 kg

Molar mass of urea = 60.06 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of solvent in Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = ?

$\Delta T^o$ = freezing point of solvent = $-10.1^oC$

i = Van't Hoff factor = 1 (for urea non-electrolyte)

$K_f$ = freezing point constant = $5.32^oC/m$

m = molality

Now put all the given values in this formula, we get

$-10.1^oC-T_s=1\times (5.32^oC/m)\times \frac{29.82g}{60.06g/mol\times 0.500kg}$

$T_s=-15.4^oC$

Therefore, the freezing point of a solution is $-15.4^oC$

5 0

3 years ago

What is the number of moles of solute in 250 mL of a .80M solution

Illusion [34]

Answer:

There are 0.1 moles of solute in 250 mL of 0.4 M solution

Explanation:

because it is

8 0

3 years ago

Calculate the concentration of oxalate ion (c2o42–) in a 0.175 m solution of oxalic acid (c2h2o4). [for oxalic acid, ka1 = 6.5 ×

Darina [25.2K]

C2H2O4 C2HO4- + H+

0.175 - x x x + y

C2HO4- c2o42– + H+

x - y y x + y

K2 = (y) (x +y) / (x-y)

As, y << x

So, K2 = (y) (x) / (x)

K2 = y =6.1 × 10^–5

Hence, concentration of (c2o42–) 6.1 × 10^–5 M

7 0

4 years ago