Ethers have a tetrahedral geometry i.e. oxygen is sp
3
hybridized. The C−O−C bond angle is 110
o
. Because of the greater electronegativity of O than C, the C−O bonds are slightly polar & are inclined to each other at an angle of 110
o
, resulting in a net dipole moment. This bond angle greater than that of tetrahedral bond angle of 109
o
28
′
. This is due to the fact that internal repulsion by the hydrocarbon part is greater than the external repulsion of the lone pair of oxygen.
Answer:
a) 0.525 mol
b) 0.525 mol
c) 0.236 mol
Explanation:
The combustion reactions (partial and total) will be:
C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
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2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O
It means that the reaction will form 50% of each gas.
a) 0.525 mol of CO
b) 0.525 mol of CO₂
c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol
So, the number of moles is the mass divided by the molar mass:
n = 11.5/100 = 0.115 mol
For the stoichiometry:
2 mol of C₇H₁₆ -------------- (37/2) mol of O₂
0.115 mol of C₇H₁₆ --------- x
By a simple direct three rule:
2x = 2.1275
x = 1.064 mol of O₂
Which is the moles of oxygen that reacts, so are leftover:
1.3 - 1.064 = 0.236 mol of O₂
Answer:
[H₂] = 13 M
[N₂] = 12 M
Explanation:
Let´s consider the following reaction at equilibrium.
3 H₂(g) + N₂(g) ⇄ 2 NH₃(g)
To find out the initial concentrations we will use an ICE chart. We recognize 3 stages: Initial, Change and Equilibrium and complete each row with the concentration or change in concentration. We will use letters for the unknown data.
3 H₂(g) + N₂(g) ⇄ 2 NH₃(g)
I a b 0
C -3x -x +2x
E a - 3x b - x 2x
We know that,
[NH₃] = 2x = 5.1 M ⇒ x = 2.6 M
Then,
[H₂] = 5.6 M = a - 3x ⇒ a = 13 M
[N₂] = 9.3 M = b - x ⇒ b = 12 M
50 percent
there are 8 atoms in this molecule
and 4 of them are oxygen molc. so:
good luck
