Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
Answer:
The volume is increased.
Explanation:
According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;
V₁ / T₁ = V₂ / T₂ -----(1)
In statement the data given is,
T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C
T₂ = 20 °C = 293.15 K
So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,
V₁ = 100 L
V₂ = Unknown
Now, we will arrange equation 1 for V₂ as,
V₂ = V₁ × T₂ / T₁
Putting values,
V₂ = 100 L × 293.15 K / 283.15 K
V₂ = 103.52 L
Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.
Answer:
-125 kJ
Explanation:
You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.
H₂C=CH₂ + H₂ ⟶ H₃C-CH₃
Bonds: 4C-H + 1C=C 1H-H 6C-H + 1C-C
D/kJ·mol⁻¹: 413 612 436 413 347
The formula relating ΔHrxn and bond dissociation energies (D) is
ΔHrxn = Σ(Dreactants) – Σ(Dproducts)
(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)
<em>For the reactant</em>s:
Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ
<em>For the products:</em>
Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ
<em>For the system</em>
:
ΔHrxn = 2700 - 2825 = -125 kJ
Answer:
The correct answer is option B.
Explanation:
Moles of 
= 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles
Then ,48 moles of NaOH will reacts with:
of
Then ,40 moles of 
will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the is an excessive reagent.
Answer:
The answer is
<h2>3.68 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass of substance = 12.50 g
volume = 3.4 mL
The density of the substance is
We have the final answer as
<h3>3.68 g/mL</h3>
Hope this helps you
