Answer: 
.
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
(1)
![S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ](https://tex.z-dn.net/?f=S_%7Bmonoclinic%7D%2BO_2%28g%29%5Crightarrow%20SO_2%28g%29%2Ftex%5D%20%5Btex%5D%5CDelta%20H%5E0_2%3D-296.36kJ)
(2)
The final reaction is:
(3)
By subtracting (1) and (2)
Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ
I think the correct answer from the choices listed above is the first option. It would be several gold atoms that will form <span>a metallic bond. It is the most logical choice since it is a metal while the others are nonmetal and a combination of metal and nonmetal. Hope this answers the question. Have a nice day.</span>
Answer:
- Partial pressure He = 276 torr
- Partial pressure Ar = 457 torr
- Total pressure = 733 torr
Explanation:
Assuming temperature remains constant, we can use Boyle's law to solve this problem: P₁V₁=P₂V₂.
Once the two flasks are connected and the stopock opened, the total volume is:
Now we use Boyle's law <em>twice</em>, to <u>calculate the new pressure of </u><em><u>each</u></em><u> gas</u>:
- He ⇒ 752 torr * 275 mL = P₂He * 750 mL
P₂He = 276 torr
- Ar ⇒ 722 torr * 475 mL = P₂Ar * 750 mL
P₂Ar = 457 torr
Finally we <u>calculate the total pressure</u>, adding the partial pressures:
- Total pressure = P₂He + P₂Ar = 733 torr
The valence electrons determine how reactive an element is, and since Sodium and other alkali metals could easily lose an electron, they become very reactive to be stable.
D because 0 is always ignorant in significant figures. hope that helps:)
