Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 100 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 10 grams of C is formed in 7 minutes. How much is formed in 28 minutes? (Round your answer to one decimal place.) grams What is the limiting amount of C after a long time? grams How much of chemicals A and B remains after a long time? A grams B grams At what time is chemical C half-formed? t = min

Answers

Answer 1

Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.  

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's  y = C

y = 3x

Still not converted sum of reaction  

for A: 100 - 2x

for B: 50 - x

Shift of x over time  

[tex]\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}[/tex]

Integration of x as regards t  

[tex]\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\[/tex]

D is the constant of integration

initial conditions: t = 0, x = 0

[tex]\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\[/tex]

hence we get:

[tex]\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}[/tex]

after t = 7 minutes , [tex]C = 10 \ g = 3x[/tex]

[tex]3x = 10\\\\x = \frac{10}{3}[/tex]

Insert the above value x into [tex]\frac{1}{(100-2x)}[/tex] equation [tex]= kt + \frac{1}{100}[/tex] to get k.  

[tex]\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\[/tex]

[tex]\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091[/tex]

therefore plugging in the equation the above value of k  

[tex]\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}[/tex]

Let y = C

, calculate C:

y = 3x

[tex]y =3 \times \frac{-45.5t}{1+0.91t}[/tex]

amount of C formed in 28 mins

[tex]x = \frac{-45.5t}{1+0.91t} ,[/tex] plug t = 28

[tex]\to x = \frac{-1274}{1+25.48} \\\\\to x = \frac{-1274}{26.48} \\\\\to x= -48.26[/tex]

therefore amount of C formed in 28 minutes is = 3x = 144.78 grams

C: [tex]y =3 \times \frac{-45.5t}{1+0.91t}[/tex]

y= 136.5 =137


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Answers

Answer:

Explanation:

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Answer:

The sum of potential energy and macroscopic kinetic energy is called mechanical energy and stays constant for a system when there are only conservative forces (no non-conservative forces). The more mass a moving object has, the more kinetic energy it will possess at the same speed.

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Answers

Answer:

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Answers

Answer:

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13)223.15 kelvin

14)363.15 kelvin

15)253.15 kelvin

Explanation:

thank u!!!

Answer:

12) 273.15

13) 223.15

14) 363.15

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Answer:

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Explanation:

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Answer:

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Answers

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18) °Celsius

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Hi :)

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Answers

Answer:

The answer is 1.50 moles

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

[tex]n = \frac{N}{L} \\ [/tex]

where n is the number of moles

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6.02 × 10²³ entities

From the question we have

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1.50 moles

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Answer: 1.50 moles

Explanation:

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Answers

Answer:

By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.

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Answers

Answer:

Solid

1. Stone

2. Brick

Liquid

1. Water

2. Oil

Gas

1. Oxygen

2. Nitrogen

Answer:

rock and sand, water and steam, smoke and hydrogen

Explanation:

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Answers

Answer:

Aluminum (Al)

Explanation:

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what are the 3 subatomic particles of the atom? What are their corresponding charges and locations within the atom?

Answers

Answer:

Protons, neutrons, and electrons are the three main subatomic particles found in an atom.

Explanation:

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Answers

Answer:

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Answers

Answer:

no

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O two s orbitals and two p orbitals.

Answers

Answer:

One s orbital and One p orbital

Explanation:

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calcium

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Answers

Strontium has an electron configuration of [Kr]5s²

Further explanation  

In an atom there are levels of energy in the shell and sub shell  

This energy level is expressed in the form of electron configurations.  

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d and f. The maximum number of electrons for each sub shell is  

s: 2 electrons  

p: 6 electrons  

d: 10 electrons and  

f: 14 electrons  

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Answer:

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