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3 years ago

What are the similarities between DIFFUSION, FACILITATED DIFFUSION and OSMOSIS?

Chemistry

2 answers:

sattari [20]3 years ago

8 0

Answer:

1)Both osmosis and diffusion equalises the concentration of two solutions.

2)Both are passive transport.

3)Do not require energy.

4)In both processes particle move or flow from the higher concentration to lower concentration until equilibrium achieved.

Thank you.

nalin [4]3 years ago

6 0

Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.

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Help help help ASFAP

gulaghasi [49]

Answer:

A directory of relatives' phone numbers

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Explanation:

5 0

3 years ago

What is the relationship among solutions solutes and solvents

jenyasd209 [6]

Answer:

Solvents are substances in which solutes dissolves while solutes are substances that dissolve in solvents and solutions result from mixing solvents and solutes.

Explanation:

A solvent is a substance such as water that dissolves a solute.
A solute is a substance that dissolves in a solvent. For example, when sodium chloride dissolves in water, sodium chloride is the solute and water is the solvent.
A solution, on the other hand, results from combining a solute and a solvent. Therefore, a mixture of water and sodium chloride forms the solution.

6 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

Archeologists can determine the age of artifacts made of wood or bone by measuring the amount of the radioactive isotope 14C pre

Ahat [919]

Answer: The age of the tool is 15539 years

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = 100

a - x = amount left after decay process = $\frac{15.5}{100}\times 100=15.5$

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{5730yr}=0.00012yr^{-1}$

b) for 15.5 % of original amount

$t=\frac{2.303}{0.00012}\log\frac{100}{15.5}$

$t=15539years$

Thus age of the tool is 15539 years

8 0

3 years ago

Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c

Alla [95]

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)= $\frac{W}{nR}$

(T₂-T₁)= $\frac{2.40 X 10^{3} }{6 X 8.314}$

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.

6 0

3 years ago