All categories

3 years ago

What’s something that describes diffusion

Chemistry

1 answer:

Mandarinka [93]3 years ago

8 0

Answer:

Diffusion is the tendency of molecules to spread out in order to occupy an available space. Gasses and molecules in a liquid have a tendency to diffuse from a more concentrated environment to a less concentrated environment. Passive transport is the diffusion of substances across a membrane.

Explanation:

here

You might be interested in

Which product is the final outcome of a nuclear generating plant? uranium heat electricity steam

weqwewe [10]

Answer:

electricity

Explanation:

i took the test and got the answer right.

8 0

3 years ago

The most common shape for a galaxy is ______.

andrew-mc [135]

The least common shape for a galaxy is cluster .

8 0

3 years ago

If a bug is traveling 5 meters across the floor in 5 seconds. How fast did it<br> travel?

Sergio039 [100]

Answer:

5 seconds

Explanation:

Speed x Time. So t=ds. t=51=5.

4 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

What would happen to other populations if they didn't have limitng factors?

Montano1993 [528]

Answer:

In the natural world, limiting factors like the availability of food, water, shelter and space can change animal and plant populations. Other limiting factors, like competition for resources, predation and disease can also impact populations. Other changes in limiting factors will cause a population to decrease.

8 0

3 years ago