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:

Explanation:
Hello!
In this case, since the energy involved during a heating process is shown below:

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

Thus, by plugging in, we obtain:

Best regards!
Actually, no. While their mass may be the same (1kg), the volume of lead is a lot smaller than that of feathers. As there is the same mass stuffed in a smaller space, it must be denser. The density of water is 1 g/cm3, so if the density of the lead is more than 1g/cm3, it has to sink
It smells so bad that no one or animal will get near it
If the grade of the ore is 37.3% nickel, then the unknown quantity to get 10 grams of nickel is 0.373 x = 10 grams or x = 10/0.373=26.8 grams or 0.0268 kg needed to dig up to recover the 10 grams of nickel. At this grade of ore, 1 kilogram would yield 373 grams of nickel.