Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Air is a mixture, Its constituens can be seperated
Higher concentration of reactants equals faster rate of reaction. Reactions occur when particles collide effectively, and by increasing the concentration of reactants, you increase the number of effective collisions, thereby making the reaction occur faster.
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:
I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
