The balanced chemical
reaction will be:
CH4 + 2O2 → CO2 + 2H2O
We are given the amount of carbon dioxide to produce from the reaction.
This will be our starting point.
560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) (
22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>
Based on Heisenberg's uncertainty principle, the position and velocity of a particle cannot be determined simultaneously with accuracy.
In other words, Heisenberg's uncertainty principle states that the more accurately we know the position of a particle the less accurately we can know its velocity. Mathematically it is given as:
Δx.mΔv >= h/2π
where: Δx = uncertainty in position
m = mass
Δv = uncertainty in velocity
h = plancks constant
Answer:
a. H2S(g)/t = 1.48 mol/s
CS2(g)/t = 0.740mol/s
H2(g)/t = 2.96mol/s
b.
Ptot /t = 981torr/min
Explanation:
a. Based on the reaction:
CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)
<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>
<em />
If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s
CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s
The H2 is produced four times faster than CH4 is decreased, that is:
0.740mol/s * 4 = 2.96mol/s
b. With the reaction:
2 NH3(g) → N2(g) + 3 H2(g)
2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.
That means 2 moles of gas are consumed and 4 moles of gas are produced.
If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =
654torr/min
The rate of change of the total pressure is rate of reactants + rate of products:
654torr/min + 327torr/min =
981torr/min
The electron geometry is tetrahedral and the molecular geometry is tetrahedral. If a molecule of CH3OCH3 is to be drawn, the two carbons would have four single bonds and the middle O would have two single bonds and the two lone pairs. Molecular geometry does not consider the lone pairs as bonds like in electron domain geometry. However, since the carbons do not contain any lone pairs, its electrons domain and molecular geometry will be the same. Therefore; Both carbons are tetrahedral for electron domain geometry and molecular geometries. The O is tetrahedral for the electron domain geometry and bent for molecular geometry.
