Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.
Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."
MH₂CO₃: (1g×2) + 12g + (16g×3) = 62 g/mol
1 mol --- 62g
0,8 mol -- X
X = 0,8×62
X = 49,6g
Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:

= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
What's the answer? It asked to be 20 characters long so just writing this.
Answer: The concentration of excess
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of
give = 1 mole of 
Thus 0.019 moles of
give = 0.019 mole of 
2. moles of 
According to stoichiometry:
1 mole of
gives = 2 moles of 
Thus 0.012 moles of
give =
moles of 

As 1 mole of
neutralize 1 mole of 
0.019 mole of
will neutralize 0.019 mole of 
Thus (0.024-0.019)= 0.005 moles of
will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of
in solution is 0.017 M.