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Mkey [24]
3 years ago
6

A student performs a titration on 25.0 mL of 0.500M HCL(aq) which is placed in an Erlenmeyer flask. Two drops of indicator are a

dded to the flask. An unknown concentration of NaOH(aq) is added to the buret the initial reading is 5.00 mL. The NaOH is added dropwise until the HCL solution is observed to be pink. The NaOH buret reading a t this point is 24.64mL. Please calculate the concentraion of the unknown NaOH Base
Chemistry
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

The unknown NaOH base has a concentration of 0.636M

Explanation:

<u>Step 1:</u> the balanced equation

NaOH + HCl → NaCl + H2O

This means for 1 mole of NaOH consumed there is 1 mole of HCl needed to produce 1 mole of NaCl and 1 mole of H2O

<u>Step 2</u>: Calculate moles of HCl used

Number of moles = Concentration *  volume =  0.5M * 25*10^-3 L =0.0125 moles

<u>Step 3</u>: Calculate moles of NaOH

Since the mole ratio for HCl and NaOH is 1:1 this means we have 0.0125 moles of NaOH for 0.0125 moles of HCl

<u>Step 4:</u> Calculate Concentration of  the unknown NaOH base

Concentration = Number of moles / Volume

Volume of NaOH = 24.64-5 =19.64 mL = 0.01964 L

Concentration = 0.0125/0.01964 = 0.636 M

The unknown NaOH base has a concentration of 0.636M

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It is a physical change.

Explanation:

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3 years ago
20. What volume of 0.350M KMnO4 solution must be diluted to prepare 600. mL of
Dafna1 [17]

Answer:

25.7 mL

Explanation:

Step 1: Given data

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  • Final volume (V₂): 600 mL
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Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

V₁ = 25.7 mL

3 0
2 years ago
Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?
AleksandrR [38]

This reaction would give rise to two products.

  • 2-bromo-3-methylhexane, and
  • 3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton \text{H}^{+} and a bromide ion \text{Br}^{-}.

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

  • \phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3
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Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon.   The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.

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2 years ago
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Final temperature of platinum :-   36.5°C

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3 0
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sineoko [7]
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3 0
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