Answer:
B, C
Explanation:
The atoms or ions with the valid Lewis dot structures are B and C.
In A;
The Lewis structure of the carbon is correct. Each of the four dots represent the four valence electrons.
The nitrogen with one dot on top, left and to the bottom and has a charge of minus 3 is wrong. For it to have a charge of -3 it must have 8 lewis dots ( two on the top, right, bottom and to the left)
The nitrogen with four dots (on top, right, bottom and to the left) is wrong.
In B;
An oxygen has two dots on top and bottom and one dot to the left and to the right. This is correct , the 6 dots represent the valence electrons of oxygen.
In C;
A carbon has two dots on top, right, bottom and to the left and a charge of plus four. This is correct because the charge indicates that it has gained four extra electrons so its valence electrons is now 8.
In D;
An oxygen has two dots on top, left and to the bottom and a charge of minus 2. This is wrong because the lewis dots are incomplete. Two dots are missing.
Answer: 103.8 g of 
, 0.65 moles of
Explanation:
To calculate the moles :
Thus 
is the limiting reagent as it limits the formation of product and 
is the excess reagent.
According to stoichiometry :
3 moles of produce = 2 moles of
Thus 0.975 moles of will produce=
of
Mass of 
Thus 103.8 g of will be produced.
Answer:
2-Butanol is the organic compound of ch3chohch2ch3
Answer is: 73.52 kJ<span> of energy is required to vaporize butane.
</span>m(C₄H₁₀) = 185 g.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 185 g ÷ 58.12 g/mol.
n(C₄H₁₀) = 3.18 mol; amount of butane.
Hvap = 23.1 kJ/mol; <span>the heat of vaporization for butane.
</span>Q = Hvap · n(C₄H₁₀).
Q = 23.1 kJ/mol · 3.18 mol; energy.
Q = 73.52 kJ.
