Answer:
where are those two images which you have sent
Answer:
The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.
Explanation:
Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;
P ∝ V
P V = K
P1 V1 = P2 V2
Parameters :
P1 = Initial pressure of the gas = 750 mmHg
V1 = Initial pressure of the gas = 2. 00 Liters
P2 = Final pressure of the gas = 600 mmHg
V2 = Fimal volume of the gas = ? Liters
Calculations :
V2 = P1 V1 ÷ P2
V2= 750 × 2. 00 ÷ 600
V2 = 1500 ÷ 600
V2 = 2.5 Liters.
Therefore, the new volume of the gas is 2. 5 Liters.
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
The statement about electronegativity that is correct is <span>D. Noble gases have the highest electronegativity values.</span>
A mixture called a solution, I think.
