<span>Hess' Law states that the enthalpy change in a reaction can be calculated from the enthalpy changes of reactions that, when combined, result in the desired reaction.
For example, to check the enthalpy change that occurs when benzene undergoes incomplete combustion to water and carbon monoxide is not an easy task, because the products invariably contain CO2. However, by combining the reactions of the complete combustion of benzene and the combustion of CO, you can get the reaction you want.
Reaction wanted: 2C6H6 + 9O2 → 12CO + 6H2O
Reactions provided: 2C6H6 + 15O2 → 12CO2 + 6H2O and 2CO + O2 → 2CO2, and their associated ΔH.
Rearrange the reactions so that, when they add up, they result in the wanted reaction.
2C6H6 + 15O2 → 12CO2 + 6H2O (leave as is; no changes to ΔH)
12CO2 → 12CO + 6O2 (reverse and multiply by 6; this changes the sign of ΔH and multiplies it by 6)
Added up, it will result in 2C6H6 + 9O2 → 12CO + 6H2O. Add up the ΔH values for the rearranged reactions to find ΔH for this particular reaction.</span>
Answer:
a,morunoxidice
Explanation:
carbon - tranite>plangose ,finite, finite,MARETE
Mol= mass (grams) /Mr
Mr of Sulfuric Acid (H2SO4): 98
mol= 329/98
=3.36 moles
Answer:
c
Explanation:
all the atoms must be balanced.
Answer:
[K⁺] = 0.107 M
[OH⁻] = 1.13 × 10⁻⁹ M
Explanation:
600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻
⇒ 0.45 × 600 × 10⁻³
= 0.27 moles of Cu²⁺ and (NO₃)²⁻
450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻
⇒ 0.25 × 450 × 10⁻³
= 0.1125 moles of K⁺ and OH⁻
Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)
Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625
=0.21375 moles which is equal to :
⇒ 0.21375/(( 600+450))× 10⁻³
= 0.21375/1050 × 10⁻³
= 0.20357 M
Given that :
(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)
We know that , Ksp = [Cu²⁺][OH⁻]²
2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²
[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357
[OH⁻] = 1.13 × 10⁻⁹ M
[K⁺] = moles of K⁺ /total volume
[K⁺] = 0.1125 / 1050 × 10⁻³
[K⁺] = 0.107 M
