All categories

3 years ago

What happens when the thermal energy of a substance increases?

Chemistry

2 answers:

MariettaO [177]3 years ago

8 0

Answer:

Explanation:

Both motion of particles and temperature increase

ohaa [14]3 years ago

4 0

Answer:

Explanation:

D: Both the motion of the particles in the substance and the temperature of the substance increase.

<em>I Hope That This Helps!!!</em>

You might be interested in

_____ stars can be between one thousand times dimmer and one million times brighter than the Sun.

dmitriy555 [2]

Answer:

The answer is

D. Main sequence

Explanation:

Main sequence stars can be between one thousand times dimmer and one million times brighter than the Sun.

What is main sequence stars?

A main sequence star is any star that is fusing hydrogen in its core and has a stable balance of outward pressure from core nuclear fusion and gravitational forces pushing inward.

6 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

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

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

4 years ago

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602

hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as $1.602 \times 10^{-19} C$ . Hence, we will calculate its number of moles as follows.