Answer:
Substitution and Na(No3) + Fe
Explanation:
Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
===============
Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
<h2>
Hello!</h2>
The answer is:
The new volume will be 1 L.

<h2>
Why?</h2>
To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.
To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.
Boyle's Law equation is:

So, we are given the information:

Then, isolating the new volume and substituting into the equation, we have:



Hence, the new volume will be 1 L.

Have a nice day!
Answer:
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)
Let's apply the thermodynamic formula to calculate the ΔG
ΔG = ΔG° + R .T . lnQ
We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q
How can we know Q? By the partial pressures (Qp)
P NO = 0.450atm
PO₂ = 0.1 atm
PNO₂ = 0.650 atm
Qp = [NO₂]² / [NO]² . [O₂]
Qp = 0.650² / 0.450² . 0.1 = 20.86
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
C6H12O6
That is the molecular formula of glucose