I'm not completely sure but if I did know I would definitely tell u
The answer is: II.The endpoint is recorded when the solution is dark red in color rather than light pink.
The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.
Phenolphthalein is colorless in acidic solutions and pink in basic solutions. If this indicator change color to dark red, more base is added and endpoint is not accurate.
If the the acid is spilled before titration, that does not make endpoint wrong and molar mass can be calculated.
In this example we can take acetic acid as carboxylic acid; basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).
Balanced chemical reaction of acetic acid and sodium hydroxide:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).
Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.
Answer:
Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. ... So, Maxwell's equations do say that light is a wave.
Explanation:
Hope this helps
99% sure its false
its arranged by atomic number now i believe
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.