Answer: Reaction 1 is non spontaneous.
Explanation:
According to Gibb's equation:
= Gibbs free energy
= enthalpy change
= entropy change
T = temperature in Kelvin
When = +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
For the given reaction 1: 
As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.
35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22
Average atomic mass closest to 35.22 amu.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Cadmium chloride is a highly soluble compound. The equation for its dissolution is:
CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)
This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
