What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?

Answer:

A

Explanation:

Nuclear fusion is a much more powerful reaction than any of the others.

Nuclear fusion of two hydrogen isotopes would expect to produce the most energy. Therefore, option (A) is correct.

Nuclear fusion can be defined as when two or more nuclei fuse together to form a single heavier nucleus. In this reaction, the matter is generally not conserved due to some of the mass of the nuclei being converted to energy.

For nuclear fusion, the nuclei must be brought closer to the reaction occurring. Each star in the universe, including sun, is shining due to nuclear fusion reactions. They produce large amounts of energy in this process.

In the Sun's core, hydrogen nuclei are get transformed into helium nuclei which are known as nuclear fusion. It takes place when four hydrogen nuclei fuse into each helium nucleus. During the Nuclear fusion process, some of the mass of hydrogen nuclei is transformed into energy.

Therefore, nuclear fusion is where hydrogen atoms combined to form helium to produce the most energy.

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Answer:

23 * 10^ 9 atoms of oxygen

Explanation:

From the molecular formula of dinitrogen tetraoxide;

molar mass of dinitrogen tetraoxide = 92 g/mol

hence;

92 g of dinitrogen tetraoxide contains 4 atoms of oxygen

x g of dinitrogen tetraoxide will contain 1 * 10^ 9 atoms of oxygen

x = 92 * 1 * 10^ 9/4

x =23 * 10^ 9 atoms of oxygen

Answer:

It was formed from slow cooling.

the angle of the Sun’s rays

Explanation:

Answer:

D. It was formed from slow cooling.

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

[tex]2SO_2 + O_2 \to 2SO_3[/tex]

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction [tex]2SO_2 + O_2 \to 2SO_3[/tex] = 0.8325;

If we want to find:

[tex]SO_2 + \dfrac{1}{2}O_2 \to SO_3[/tex]

Then:

[tex]K_c = (0.8325)^{1/2}[/tex]

[tex]\mathbf{K_c = 0.912}[/tex]

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

Answer:

d) The rate of forward reaction equals the rate of the reverse reaction

Explanation:

A dynamic equilibrium is attained when the rate of forward reaction is the same as reverse reaction. It occurs majorly in a reversible reaction. Thus, no change occurs in the reaction.

The process can be control by the immediate removal of the product as it is formed. This ensure that the reaction stops when all reactants are used up.

Answer:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 =  log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 =  log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 =  log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

Explanation:

Answer:

c

Explanation:

Answer:

a)Weight of Y extracted from water in single extraction = 67.2 g

b) Total weight of Y extracted from water in two successive extractions = 70.9 g

Explanation:

Distribution coefficient, Kd = Cs / Cm

where Cs  is concentration of solute in methylene chloride; Cm is concentration of solute in water

a) Let x g of solute Y be extracted into methylene chloride

72.0 - x g of Y will be left in water

Kd = (x/180 mL of methylene chloride) / (72.0 - x/180 mL of water) = 14

14 = 180 * x / 180 * 72.0 - x

x = 14(72.0 - x)

x = 1008 - 14x

15x = 1008

x = 1008/15

x = 67.2 g

Weight of Y extracted from water in single extraction = 67.2 g

b) First extraction: Let x g of solute Y be extracted into methylene chloride

72.0 - x g of Y will be left in water

14 = x/90 / 72.0 - x/180

14 = 180x/ 90(72.0 - x)

14 = 2x/ 72.0 - x

2x = 14(72.0 - x)

2x =  1008 - 14x

16x = 1008

x = 63.0 g

Second extraction: Amount of solute left in water after first extraction = 72 - 63 = 9.0 g

Let x g of solute Y be extracted into methylene chloride

9.0 - x g of Y will be left in water

14 = x/90 / 9.0 - x/180

14 = 180x/ 90(9.0 - x)

14 = 2x/ 9.0 - x

2x = 14(9.0 - x)

2x =  126 - 14x

16x = 126

x = 7.90 g

Total weight of Y extracted = (63 + 7.90) g

Total weight of Y extracted from water in two successive extractions = 70.9 g

a) The Weight of Y extracted from water in single extraction is 67.2 g

b) The Total weight of Y extracted from water in two successive extractions should be 70.9 g

Here

The distribution coefficient, Kd should be

= Cs / Cm

where

Cs  should be a concentration of solute in methylene chloride;

Cm should be a concentration of solute in water

a) Let  us assume x g of solute Y that should be extracted into methylene chloride

So,

72.0 - x g of Y will be left in water

And, now

Kd = (x/180 mL of methylene chloride) / (72.0 - x/180 mL of water) = 14

So,

14 = 180 * x / 180 * 72.0 - x

x = 14(72.0 - x)

x = 1008 - 14x

15x = 1008

x = 1008/15

x = 67.2 g

b)

Now First extraction:

Let x g of solute Y be extracted into methylene chloride

So,

72.0 - x g of Y will be left in water

14 = x/90 / 72.0 - x/180

14 = 180x/ 90(72.0 - x)

14 = 2x/ 72.0 - x

2x = 14(72.0 - x)

2x =  1008 - 14x

16x = 1008

x = 63.0 g

Second extraction:

Amount of solute left in the water after the first extraction that should be

= 72 - 63

= 9.0 g

Now

Let x g of solute Y be extracted into methylene chloride

9.0 - x g of Y will be left in water

14 = x/90 / 9.0 - x/180

14 = 180x/ 90(9.0 - x)

14 = 2x/ 9.0 - x

2x = 14(9.0 - x)

2x =  126 - 14x

16x = 126

x = 7.90 g

Total weight of Y extracted = (63 + 7.90) g

= 70.9g

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Answer:

11.023 pounds

Explanation:

Extremophiles are bacteria, archaea, as well as eukarya across all three domains of life. Cells make up every organisms.

Extremophiles contain bacteria, archaea, as well as eukarya from all three domains of life.

Any organic, biological system which works as a separate entity was referred to as an organism.

Extremophiles are organisms that can survive in harsh settings, such as those with high temperatures, radiation, salt, or pH levels. In the planet's evolutionary development, several creatures have been ecologically dominating.

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Answer:

74.3kJ are transeferred

Explanation:

In the bomb calorimeter, the burning of C₈H₁₈ is producing heat that is been absorbed for the calorimeter and the water.

Heat absorbed calorimeter:

837J/°C * (33.8°C-15.0°C) = 15735.6J

Heat absorbed water:

C*m*ΔT

C specific heat of water (4.184J/g°C), m is mass of water (745g), and ΔT change in temperature (33.8°C - 15°C = 18.8°C)

4.184J/g°C*745g*18.8°C = 58601.1J

Heat released by the reaction:

15735.6J + 58601.1J = 74336.7J

Answer:

the correct answer would be A. acidity

Answer:

The answer is A.) Acidity

Answer:

The answer is B(scale)

Explanation:

Since the question asked about a range, the scale would be the most logical answer since scales are used to measure.

Answer:

the answer is C.

Explanation:

Answer:

O The sound waves are reflected.

Explanation:

When sound waves released by a bat hits an object, the sound waves are reflected back.

Bats used sound reflection for echolocation to determine the position of nearby objects.

Also, they use it to determine the shape and size of objects.

Answer:

B

The sound waves are reflected.

Explanation:

Answer:

The answer is 16

Explanation:

1) Both cells have a cell membrane.

2) A nucleus is present in only Eukaryote cells.

3) I'm pretty sure Prokaryote cells are bacteria.

Hope this helps!! :)

Answer:

THe sun evapored the water and the wind helped dry it to.

Explanation:

Answer:

L× W×H

Answer:

The correct answer is L x W x H :)

Answer:

red

Explanation:

blue litmus paper turns red under acetic conditions and red lettuce paper turns blue under basic and alkaline conditions with the color change occurring over the PH range the neutral litmus paper is purple

The answer is red.

took the test.

Answer:

The answer is "Option a".

Explanation:

Multiply the mass of the solutions by 3.17% to get the mass of [tex]BaCl_2[/tex]:

[tex]\to 427 \ g \times 0.0317 = 13.56\ g[/tex]

Divide this mass with BaCl2's molecular concentration to acquire the mole ratio:  

[tex]molar mass = 208.24 \frac{g}{mol}\\\\moles = \frac{13.56}{208.24} = 0.065 \ moles \\\\[/tex]

                       [tex]= 6.5 \times 10^{-2} \ moles[/tex]

A 427 g sample is 3.17% by mass barium chloride contains 6.5 × 10⁻² moles of BaCl₂.

A 427 g sample is 3.17% by mass barium chloride, that is, there are 3.17 g of barium chloride per 100 g of sample.

The mass of barium chloride in 427 g of the sample is:

[tex]427gSample \times \frac{3.17gBaCl_2}{100gSample} = 13.5gBaCl_2[/tex]

We can calculate the number of moles in 13.5 g of barium chloride using its molar mass (208.23 g/mol).

[tex]13.5 g \times \frac{1mol}{208.23g} = 6.5 \times 10^{-2} mol[/tex]

A 427 g sample is 3.17% by mass barium chloride contains 6.5 × 10⁻² moles of BaCl₂.

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Answer:

The ball's kinetic energy is transforming to potential energy.

Explanation:

the ball is stopping resulting in potential energy

Answer:

- 0.454 °C

Explanation:

From the given information:

The mass of KCl = 10 grams

The molar mass of KCl = 74.55 g/mol

The molality can be calculated as:

[tex]Molality \ m = \dfrac{mass \ of \ KCl \times 1000}{molar \ mass \ of \ KCl \times 1100.0}[/tex]

[tex]Molality \ m = \dfrac{10 \ g \times 1000}{74.55 \ g/mol \times 1100.0}[/tex]

Molality m = 0.1219 M

For freezing point depression;

[tex]\Delta T_R = i \times k_f \times m[/tex]

The van't Hoff's Factor (i) for KCl  = 2

The molar depression of freezing point constant [tex]k_f[/tex] = 1.86° C/m

∴

[tex]\Delta T_R = 2 \times 1.86 \times 0.1219[/tex]

[tex]\mathbf{\Delta T_R = 0.454^0\ C }[/tex]

However, the freezing point of water is known to be = 0°C

Therefore, the freezing point of the solution

= 0 °C - 0.454 °C

= - 0.454 °C

Answer:

0.133

Explanation:

reaction between KIO3 and KI in acidic medium

IO3⁻ +5I⁻ +6h⁺ → 3I₂ + 3H₂O

I₂ reacts with thiosulphate

NaS₂O₃  → 2Na⁺ + S₂O₃²⁻

net reaction

IO⁻₃ + 6H⁺ + 6S₂O₃³⁻ → I⁻ + 3S₄O₆²⁻ + 3H₂O

mole of KIO₃

= molarity x volume

[tex]\frac{0.02mol}{L} *0.01L[/tex]

= 0.00002mol

a mole of KIO₃ has reaction with 6 mol of S₂O₃²⁻

= 2x6x10⁻⁵

= 0.00012 mol

volume = 0.90 ml

1 ml = 0.001L

0.90ML  = 0.0009L

to get concentration,

molarity/volume

= 0.00012/0.0009

= 0.133m

The concentration of the standard sodium thiosulfate solution is ; 0.133

The net chemical reaction equation

IO₃⁻  +  6H⁺  +   6S₂O₃³⁻ ---->  I⁻  +  3S₄O₆²⁻  + 3H₂O

First step : Determine the moles of  KIO₃

number of moles = molarity * volume

                            = 0.02 mol / L * 0.01 L

                            = 0.00002 mol

From the net chemical reaction equation

one ( 1 ) mole of KIO₃ reacts with 6 moles of S₂O₃²⁻

∴ number of moles in the reaction = 6 * 0.00002 = 0.00012 mol.

Final step : Determine the concentration of sodium thiosulfate solution

Given Volume = 0.90 ml = 0.0009 L

∴ concentration of sodium thiosulfate solution

   =  0.00012 / 0.0009  =  0.133

Hence we can conclude that The concentration of the standard sodium thiosulfate solution is ; 0.133

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Answer:

See explanation

Explanation:

Coal is primarily used for electricity generation. The burning of coal leads to emission of gases such as oxides of carbon, oxides of sulphur, oxides of nitrogen and water vapour. All these go straight into the atmosphere.

Petroleum is primarily used as transportation fuels. The burning of petroleum oils mostly leads to the emission of oxides of carbon and sulphur together with water vapour into the atmosphere.

Answer:

21 is the max but with the bread ypu could make 22 if you had 3 more cheese

Answer:

The actual concentration of the hydrochloric acid is 1.2M

Explanation:

The formula to be used here is that of concentration (popularly used during titration);

CₐVₐ/CbVb =nₐ/nb

where Cₐ is the concentration of acid (supposed to be 1.0 M but unsure)

Vₐ is the volume of acid (100 ml)

Cb is the concentration of base (1.3 M)

Vb is the volume of base (89.13 ml)

nₐ is the volume of acid

nb is the volume of base

The equation for the reaction described in the question is

HCl + NaOH ⇒ NaCl + H₂O

we can see from the above equation the ratio of the number of moles for both the acid and the base is 1:1

Thus;

Cₐ × 100/1.3 × 89.13 =1/1

Cₐ = 1.3 × 89.13/100

Cₐ = 1.2M

The actual concentration of the hydrochloric acid is 1.2M

Answer:

The given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

The given general reaction has been:

[tex]\rm AB\;+\;CD\;\rightarrow\;AD\;+\;CB[/tex]

In the given reaction, there has presence of compound AB and CD. The reaction between the two results in the formulation of AD and CB. In the reaction, the B has been more attracted by C and forms bond with C by replacing D.

In the same way, A has been more attracted with D and forms bond with D replacing C.

The type of reaction in which the more reactive element replaces the less reactive element has been termed as replacement reaction. Thus, the given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

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Answer:

Will be more acidic

Explanation:

The equilibrium of NH3 in water is:

NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).

Where equilibrium constant, Kb, is:

Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]

From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:

1.85x10⁻⁵ = [X] [X] / [0.10M]

1.8x10⁻⁶ = X²

X = 1.34x10⁻³ = [OH⁻]

As pOH = -log[OH⁻] = 2.87

And as pH = 14 - pOH

pH of the 0.10M NH3 is 11.13

Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:

pOH = pKb + log [NH4⁺] / [NH3]

Where pKb is -log Kb = 4.74 and [] are moles of both compounds.

Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:

pOH = 4.74 + log [0.0132] / [0.0132]

pOH = 4.74

pH = 14 - 4.74 = 9.26

That means the pH of the resulting solution will be more acidic

Answer:

C

Explanation:

It will be the same. That's what moving down a column means. The properties are going to be quite similar.