Answer:
see calculations below
Explanation:
Given: HOAc ⇄ [H⁺] + [OAc⁻]
C(i) 0.12M 0M 0.15M
mix => 5.55ml(0.092M NaOH) / (50ml + 5.55ml)
= 0.00555(0.092)mole NaOH / 0.0555 L Soln
= 0.0092M in NaOH is added into the initial buffer solution
= 0.0092M in OH⁻ (NaOH is a strong base => 100% ionized)
Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer . equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by . 0.0092M each.
Therfore ...
Given: HOAc ⇄ [H⁺] + [OAc⁻]
C(i) 0.12M 0M 0.15M
ΔC - 0.0092M +x +0.0092M
C(f) 0.1108M x 0.1592M => New Concentrations . after adding 0.0092M . NaOH
Substituting new acid and ion concentrations into Ka expression ...
Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M
=> x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions
*units of Ka are Molar
FYI => Adding a strong base to a buffer solution will shift pH to more basic.
Adding a strong acid to a buffer solution will shift pH to more acidic.
=> (such is a good way to check that your buffer calculations are correct.)
NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore ...
[HOAc] = 0.1108M
[NaOAc] = 0.1592M
[NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction. Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.
Hope this helps. :-)
what are possible source of error for rusting of a nail
Answer:
A nail can rust when exposed to oxygen. the molecules of iron on the surface of the nail exchange atoms with the oxygen in the air and produce a new substance, the reddish brown ferrous oxide i.e rust.
Please give me the answer please
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
Moles CaCO₃ -Molar mass: 100.09g/mol-
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
Moles HCl:
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
Where P is pressure = 1atm assuming STP
V volume in L
n moles = 1.25x10⁻³ moles CO₂
R gas constant = 0.082atmL/molK
T = 273.15K at STP
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
A. 30cm³How does the entropy change in the reaction 2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)?
I will mark brainliest!! Thank you so much!!
Answer:
The entropy increases!!!
Explanation:
a pex
The entropy increases in the reaction.
What is entropy?Entropy is defined as the measure of the disorder of a system.Entropy is an extensive property of a thermodynamic system, to put it in simple words, its value changes depending on the amount of matter that is present.Entropy is denoted by the letter S and has units of joules per kelvin (JK−1)The entropy increases in the reaction if the total number of product molecules are greater than the total number of reactant molecules.
2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)
In the above reaction, the product molecules are greater than the reactant molecules. Hence, entropy increases.
Hence, we can conclude that option A is the answer.
To learn more about entropy here
https://brainly.com/question/22861773
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