Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have <u>8 electrons</u> (except hydrogen). Additionally, each carbon would have <u>4 valence electrons</u>, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an <u>additional carbon</u>. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have <u>more area of interaction</u> for ethane. If we have more area of interaction we have to give <u>more energy</u> to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
I hope it helps!
Answer: :)
Explanation:
In a covalent bond, the atoms bond by sharing electrons. Covalent bonds usually occur between nonmetals. For example, in water (H2O) each hydrogen (H) and oxygen (O) share a pair of electrons to make a molecule of two hydrogen atoms single bonded to a single oxygen atom
Answer: At equilibrium, the partial pressure of 
is 0.0330 atm.
Explanation:
The partial pressure of is equal to the partial pressure of 
. Hence, let us assume that x quantity of 
is decomposed and gives x quantity of and x quantity of .
Therefore, at equilibrium the species along with their partial pressures are as follows.
At equilibrium: 0.123-x x x
Now, expression for 
of this reaction is as follows.
![K_{p} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\\0.0121 = \frac{x \times x}{(0.123 - x)}\\x = 0.0330](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPCl_%7B3%7D%5D%5BCl_%7B2%7D%5D%7D%7B%5BPCl_%7B5%7D%5D%7D%5C%5C0.0121%20%3D%20%5Cfrac%7Bx%20%5Ctimes%20x%7D%7B%280.123%20-%20x%29%7D%5C%5Cx%20%3D%200.0330)
Thus, we can conclude that at equilibrium, the partial pressure of is 0.0330 atm.
Answer: 6.162g of Ag2SO4 could be formed
Explanation:
Given;
0.255 moles of AgNO3
0.155 moles of H2SO4
Balanced equation will be given as;
2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)
Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,
Therefore the number of moles of Ag2SO4 produced is given by,
n(Ag2SO4) = 0.255 mol of AgNO3 ×
[0.155mol H2SO4 ÷ 2 mol AgNO3] x
[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]
= 0.0198 mol of Ag2SO4.
mass = no of moles x molar mass
From literature, molar mass of Ag2SO4 = 311.799g/mol.
Thus,
Mass = 0.0198 x 311.799
= 6.162g
Therefore, 6.162g of Ag2SO4 could be formed
