Answer:
have you finished this yet
Explanation:
Answer:
The standard entropy or reaction ΔS° =+324.6J/mol.K
Explanation:
The reaction is
2H2O(l) → 2H2(g) + O2(g)
Standard Molar Entropies of Selected Substances at 298 K are
Substance S∘ (J/mol⋅K)
H2(g) 130.6
O2(g) 205.0
H2O(g) 188.8
H2O(l) 69.9
The standard entropy of reaction will be obtained from standard entropy of formation as given below
ΔH°= 2(130.6)+(205.0)-2(69.9) (Jol/mol.K)
ΔH°=261.2+205.0-139.8
ΔH°=+324.6Jol/mol.K
Answer: The balloon will not survive.
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 1 atm
= final pressure of gas = 0.05 atm
= initial volume of gas = 4.0 L
= final volume of gas = ?
= initial temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
= final temperature of gas = ![-75^oC=273-75=198K](https://tex.z-dn.net/?f=-75%5EoC%3D273-75%3D198K)
Now put all the given values in the above equation, we get:
![\frac{1\times 4.0}{298}=\frac{0.05\times V_2}{198}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5Ctimes%204.0%7D%7B298%7D%3D%5Cfrac%7B0.05%5Ctimes%20V_2%7D%7B198%7D)
![V_2=53.2L](https://tex.z-dn.net/?f=V_2%3D53.2L)
As the balloon has the capacity to expand to a 50 liter volume before bursting, it can not survive as it occupies volume higher than 50 L
Answer: it's nonmetal
Explanation: Sulfur has characteristics of nonmetals. It does not have luster, it cannot conduct electricity, and it is brittle. On the periodic table, sulfur is in group 16/VIA, in the third period. The group is called the oxygen group or the chalcogens.