All categories

2 years ago

MARKING BRAINLEIST + 20 POINTS Explain why energy must be added to a solid to cause it to melt?

1 answer:

4 0

Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart so that the molecules can move around at comparable kinetic energies

You might be interested in

Water has a higher boiling point than acetone does. Which of the following statements about water and acetone in the liquid stat

Semenov [28]

Answer:

Water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.

Explanation:

Hello.

In this case, since no options are given we can infer from the statement that due to water's higher boiling point than acetone we can conclude that when they are in liquid state, water has stronger intermolecular forces which allow its particles to be held in a stronger way in comparison to the acetone's molecules, for that reason, more energy will be required in order to separate them and promote the boiling process, which is attained via increasing the temperature. Besides, less energy will be required for the separation of the acetone's molecules in order to boil it when liquid, therefore, a lower temperature is required.

In such a way, we can sum up that water's boiling point is higher than acetone's one due to the stronger intermolecular forces it has in liquid phase.

Regards.

7 0

3 years ago

A first order reaction has a rate

EleoNora [17]

Answer:

one-half

Explanation:

cuz for a first order reaction is a half life independent of concentration and constant over time

4 0

3 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

2 years ago

In the hot climate, human bodies had to adapt by increasing the number of ____________ on the skin to cool off the body.

irga5000 [103]

D none of the above

4 0

2 years ago

Read 2 more answers

At what temperature, would the volume of a gas

PtichkaEL [24]

Explanation:

P1V1 = nRT1

P2V2 = nRT2

Divide one by the other:

P1V1/P2V2 = nRT1/nRT2

From which:

P1V1/P2V2 = T1/T2

(Or P1V1 = P2V2 under isothermal conditions)

Inverting and isolating T2 (final temp)

(P2V2/P1V1)T1 = T2 (Temp in K).

Now P1/P2 = 1

V1/V2 = 1/2

T1 = 273 K, the initial temp.

Therefore, inserting these values into above:

2 x 273 K = T2 = 546 K, or 273 C.

Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.

From the ideal gas equation:

V = nRT/P or at constant pressure:

V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.

8 0

2 years ago