Particulate movement and energy increases when a liquid is heated to its boiling point. Explanation: When a liquid is heated to its boiling point, the form of the liquid changes. ... The particles with cinematic energy begin to move more randomly. So we can say that the particle movement and the energy grow.
Answer:
Option b. 22 g of He will have the greatest volume at STP
Explanation:
In order to determine the volume, we apply the Ideal Gases Law equation:
P . V = n . R . T
V = n . R . T / P
R, T and P are the same in all the situation we must define n (number of moles).
The one that has the greatest number of moles will have the greatest volume at STP
22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne
22g of He . 1mol / 4 g = 5.5 moles of He
22 g of O₂ . 1mol / 32g = 0.68 moles of O₂
22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂
Answer:
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Given: Kc = 0.50
Temperature = ![400^oC=[400+273]K=673K](https://tex.z-dn.net/?f=400%5EoC%3D%5B400%2B273%5DK%3D673K)
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(3+1) = -2
Thus, Kp is:
Answer:
Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
