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kolezko [41]
3 years ago
7

What volume of 0.117 M HCl is needed to neutralize 28.67ml of 0.137 m KOH?

Chemistry
1 answer:
shutvik [7]3 years ago
3 0
To determine the volume of the acid needed in the reaction, we first have to know the reaction involved in the process. Since it is a neutralization reaction, then it should produce a salt and water. The reaction would be:

HCl + KOH = KCl + H2O

We use this reaction to relate the substances. We calculate as follows:

0.137 M KOH (.02867 L solution ) = 3.92779x10^-3 mol KOH
3.92779x10^-3 mol KOH ( 1 mol HCl / 1 mol KOH ) = 3.92779x10^-3 mol HCl

Volume of HCl = 3.92779x10^-3 mol HCl / 0.117 M HCl 
                           = 0.03357 L HCl or 33.57 mL HCl needed
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Answer:

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6 0
4 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
Which of the following objects does not decompose?
Anettt [7]
B) Plastic Water Bottle
6 0
3 years ago
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50 POINTS! ANSWER FAST!
Korvikt [17]

<u>Given:</u>

The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J

The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J

<u>To determine:</u>

The final energy state Efinal of the electron

<u>Explanation:</u>

Since energy is being released, this suggests that Efinal < Einitial

i.e. ΔE = Einitial - Efinal

Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J

Ans: A)

The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J

6 0
3 years ago
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barxatty [35]
<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
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NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>
4 0
3 years ago
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