10. Capital C and D represent products of chemical reaction, the capital A and B represent reactants, <span>the lower case letter represent coefficients (how many atoms or molecules in chemical reaction).
12. According to </span><span>Le </span>Chatelier's principle (if<span> the concentration is changed, that will shift the equilibrium to the side that would reduce that change in concentration)</span> <span>the equilibrium shift to the left.
13. </span>According to Le Chatelier's principle the equilibrium shift to the right.
14. According to Le Chatelier's principle (<span>When the reaction is </span>exothermic<span>, heat is included as a product)</span> the equilibrium shift to the right.
9.5⋅<span>10<span>2
</span></span>Explanation:
To figure out how many atoms of copper you get in 1 gram of copper, you need to use copper's molar mass.
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Answer:
Ag 0 is the reducing agent.
Explanation:
Reducing -> gaining electrons
Oxidizing -> losing electrons
Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.
Answer: a,b
Explanation: because a scientific theory is something that been thought of and tested multiple times.
