In stars helium atoms are used with hydrogen atoms in order to produce energy.
Answer: A )
Answer:
Melting of snow
Evaporation of water from desk
Explanation:
Processes that increase the entropy of the universe are those processes that have an increased disorderliness. We should note that there are three principal states of matter which are the liquid, gas and solid. The gaseous state is the most disorderly while the solid is the least disorderly.
Now. We can see that the cooling of a hot cup of coffee is a process that needs or leads to a loss in temperature which obviously decreases disorderliness of the universe.
The melting of snow however is a process that leads to an increase in the disorderliness of the universe. It entails moving from the solid state to the liquid state. It tends to move to a more disordered state indicating an increase in the entropy of the universe.
The evaporation of water from the desk is quite similar to that above. Hence since we are moving from the liquid to the gaseous state via evaporation, we can state that the entropy of the universe has increased since we have moved from a state with a lesser degree of disorderliness to a state that is more disordered I.e from liquid to gaseous state.
Answer:
HF has the higher boiling point because HF molecules are more polar. Part B: CHBr3 molecules possess stronger intermolecular interaction due to higher molar mass than CHCl3
Explanation:
Fluorine is more electronegative than chlorine. This implies that HF is more polar and possess stronger hydrogen bonds than HCl molecules.
In part B, the magnitude of dispersion forces depend on molar mass, the greater the molar mass, the greater the magnitude of dispersion forces between molecules, hence CHBr3 has a greater boiling point than CHCl3
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In

, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.