05.04.2021

A powder sample weighing 250 g and having a surface area of 1 m2/g is dispersed in 1 L of water. After 10 minutes, 0.4 g of the sample is dissolved. The aqueous solubility of the sample is 0.01 g/cm3. a) Calculate k. Assume that sink conditions are prevailing and that the surface area of the sample is constant. b) How long will the sink conditions prevail

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24.06.2023, solved by verified expert
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a)

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08

b)

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 hours

Explanation:

Given

Surface Area (S) of 250 grams of powder sample A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 square meter.

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 square centimeter

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 grams/minute

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 mg per second

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 gram per cubic meter A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 mg/ml

a)

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 cm /sec

b) Time taken

Time taken to sink

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08

Time in hours

A powder sample weighing 250 g and having a surface, №17887147, 05.04.2021 18:08 hours

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Chemistry
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a)

K = 2.64 * 10^{-8}

b)

15 \frac{5}{8} hours

Explanation:

Given

Surface Area (S) of 250 grams of powder sample = 250 square meter.

S = 2500000 square centimeter

V\frac{dC}{dT} = 0.40 grams/minute

V\frac{dC}{dT} = \frac{dM}{dT} = \frac{0.40*60}{10} = 0.66 mg per second

C_s = 0.01 gram per cubic meter = 10 mg/ml

a)

\frac{dM}{dT} = K SC_s

0.66 =  K* 2500000* 10\\

K = 2.64 * 10^{-8} cm /sec

b) Time taken

Time taken to sink

\frac{10*60}{0.4} * 37.5  = 56,250

Time in hours

15 \frac{5}{8} hours

Chemistry
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Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
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Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the
Chemistry
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P Answered by PhD

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Chemistry
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Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N ×328 K

Solving:

Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

Chemistry
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15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Chemistry
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P Answered by PhD
Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g
Chemistry
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P Answered by PhD
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined a
Chemistry
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P Answered by PhD
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified in terms of hydrogen ions or hydroxide ions, or in terms of electron pairs. (look at the picture)
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic

But pH= - log [H^+]
pOH = -log [OH^-]
Then;
pH + pOH = 14
Hence;
pH = 14 - pOH

a. [H+] = 6.0 x 10-10M
pH= 9.22 is basic
b. [OH-] = 30 × 10-2M
pH = 13.5 is basic
C. IH+1 = 20× 10-7M
pH = 0.56 is acidic
d. [OH-] = 1.0 x 10-7M
pH = 7 is neutral
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified

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