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algol13
3 years ago
14

4. After reaching the final titration endpoint the solution will be cloudy white. As time goes on the solution will turn back to

a cloudy dark purple color. Why does this occur if you have already reached the endpoint
Chemistry
1 answer:
AlexFokin [52]3 years ago
3 0

Answer: hello some part of your question is missing below is the missing part

In an experiment to determine the % of ascorbic acid in Vitamin C Tablets by Titration with Potassium Bromate,

answer:

Oxidation half reaction of Vitamin C

Explanation:

The solution will turn cloudy dark purple even after reaching endpoint when allowed to settle with time. because of the Oxidation half reaction of Vitamin C. also during the Titration process few drops of starch solution will be added to help determine the endpoint of the experiment .

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Fudgin [204]
The potential energy the bucket started with transforms into kinetic energy as it moves.
6 0
3 years ago
Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fraction 0.78 N2, 0.21 O2, and 0.010 Ar, what i
Artist 52 [7]

Answer:

1.293 kg/m³

Explanation:

Density: this can be defined as the ratio of the mass of a body and its volume.

The S.I unit of density is kg/m³

D = m/v............................ Equation 1

D = density of air, m = mass of air, v = volume of air.

<em>Note: Since the air is made up of nitrogen, oxygen and argon. the mass and volume of air will be the sum of the mass and sum of the volume of nitrogen, oxygen, and argon respectively.</em>

(i) Mass of one mole of N₂ = 28 g

Therefore, mass of 0.78 mole of N₂ = 28×0.78 = 21.84 g.

(ii) Mass of one mole of O₂ = 32 g

Therefore, mass of 0.21 mole of O₂ = 32×0.21 = 6.72 g

(iii) Mass of one mole of Ar = 40 g

Therefore mass of 0.01 mole of Ar = 0.01×40 = 0.4 g

m = 21.84 + 6.72 + 0.4 = 28.96 g

Also Note: One mole of every gas at standard temperature and pressure = 22.4 dm³

Therefore,

(i) volume of 0.78 mole of N₂ = 22.4×0.78 = 17.47 dm³

(ii) volume of 0.21 mole of O₂ = 0.21×22.4 = 4.704 dm³

(iii) volume of 0.01 mole of Ar = 0.01×22.4 = 0.224 dm³

Therefore,

v = 17.47+4.704+0.224

v = 22.398 dm³

Substituting the value of v and m into equation 1

D = 28.96/22.398

D = 1.293 g/dm³

D = 1.293 kg/m³

5 0
4 years ago
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Explain why a climax community is not always a forest.
kaheart [24]
They are all in forest climax means forest
4 0
3 years ago
Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum metal required to react
11Alexandr11 [23.1K]

Answer:

129.4g

Explanation:

Reaction equation":

            4Al   + 3O₂  →  2Al₂O₃

Given mass of oxygen  = 115g

Unknown:

mass of aluminum metal that reacted;'

Solution:

We need to convert the given mass of O₂ to mole;

  Number of moles  = \frac{mass}{molar mass}

       Molar mass of O₂ = 2(16) = 32g/mol

    Number of moles  = \frac{115}{32}   = 3.6mol

From the balanced reaction equation:

  3 moles of O₂ reacted with 4 mole of Al

  3.6 moles of O₂ will react with \frac{3.6 x 4}{3}   = 4.8mol

Mass of Al  = number of moles x molar mass

Mass of Al  = 4.8 x 27  = 129.4g

7 0
3 years ago
How many moles are in silicon dioxide if the mass is 3.4x10-7
Kisachek [45]

Answer:

28 as compare to ul formula

3 0
3 years ago
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