Answer:
∴ΔH₂ = - 12,258 KJ
Explanation:
Enthalpy:
Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.
The heat absorbes or releases in a closed system is the change of enthalpy of the system.
Given reactions are:
Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O, ΔH₁= - 2043 KJ
Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O, ΔH₂=?
Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈ is combusted in reaction 2.
We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)
= - 12,258 KJ
∴ΔH₂ = - 12,258 KJ
The concentration of [H3O+] will be 6.3 x 
M
<h3>pH</h3>
Mathematically, pH = -log [H+] or -log [H3O+]
With a pH of 13.2:
-log [H3O+] = 13.2
log [H3O+] = -13.2
[H3O+] = 6.3 x M
More on pH can be found here: brainly.com/question/491373
#SPJ1
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Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
