Data:
M (Molar Mass) of C2H6
C = 2*12 = 24 amu
H = 6*1 = 6 amu
---------------------
MM C2H6 = 24+6 = 30 g/mol
P (pressure) = 1.6 atm
V (volume) = 12.7 L
<span>R = 0,082 atm .L/mol.K
n (</span>Number of mols) →
![n = \frac{m}{MM} m (mass) = ? T = 24ºC Celsius to Kelvin TK = TºC + 273 TK = 24 + 273 TK = 297 By the equation of state of the gases or equation of Clapeyron, we have: [tex]P*V = n*R*T](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Bm%7D%7BMM%7D%20m%20%28mass%29%20%3D%20%3F%20T%20%3D%2024%C2%BAC%20%20Celsius%20to%20Kelvin%20TK%20%3D%20T%C2%BAC%20%2B%20273%20TK%20%3D%2024%20%2B%20273%20TK%20%3D%20297%20%20%20%3Cspan%3EBy%20the%20equation%20of%20state%20of%20the%20gases%20or%20equation%20of%20Clapeyron%2C%20we%20have%3A%0A%20%20%20%3C%2Fspan%3E%5Btex%5DP%2AV%20%3D%20n%2AR%2AT%20)
<span>
Since </span>
<span>, we can perform the following substitution in the above Clapeyron equation:
</span>
multiply cross
Solving:
Answer:
25.03 grams of ethane gas
Because their electronic configurations have the same number of electrons in the outermost shell.
G hi m m be DChjjbvccvbnjgf
Answer:
15.14×10⁷ g
Explanation:
Given data:
Number of particles of NaCl = 1.56×10³⁰ particles
Mass of sodium chloride = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ particles
1.56×10³⁰ particles × 1 mol / 6.022 × 10²³ particles
0.259 ×10⁷ mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.259 ×10⁷ mol × 58.44 g/mol
Mass = 15.14×10⁷ g
<u>Answer:</u> The enthalpy of the overall chemical equation is -205.7 kJ
<u>Explanation:</u>
Hess's law is defined as the law which states that the total amount of heat absorbed or released for a given chemical reaction will remain the same whether the process occurs in one step or several steps.
In this law, a chemical equation is treated as an ordinary algebraic expression that can be added or subtracted to yield the required overall equation.
The enthalpy change or heat of the overall reaction is the summation of the enthalpies or heats of the intermediate reactions involved in the process.
The given overall chemical equation follows:
The intermediate equations for the above reaction are:
(1) 
(2) 
(3) 
According to Hess law, the enthalpy of the reaction becomes:
![\Delta H^o_{rxn}=[1\times (\Delta H_1)] + [1\times (-\Delta H_2)] + [2\times (\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%28%5CDelta%20H_1%29%5D%20%2B%20%5B1%5Ctimes%20%28-%5CDelta%20H_2%29%5D%20%2B%20%5B2%5Ctimes%20%28%5CDelta%20H_3%29%5D)
Putting values in above expression, we get:
![\Delta H^o_{rxn}=[1\times (74.6)] + [1\times (-95.7)] + [2\times (-92.3)]\\\\\Delta H^o_{rxn}=-205.7kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%2874.6%29%5D%20%2B%20%5B1%5Ctimes%20%28-95.7%29%5D%20%2B%20%5B2%5Ctimes%20%28-92.3%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-205.7kJ)
Hence, the enthalpy of the overall chemical equation is -205.7 kJ
