Complete and balance the nuclear equations for the following fission reactions.
a. 23592U+10n→16062Sm+7230Zn+?10n. Express your answer as a nuclear equation.

b. 23994Pu+10n→14458Ce+?+210n

Answer:

pH = 12.43

Explanation:

...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

Moles KOH:

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

Moles HN3:

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

Answer:

Explanation:

This apparent disparity of the free energy can be taken into account because:

the free energy produced by the hydrolysis of one ATP is adequate enough under psychological circumstances.

The Na-K ATPase aids the pumping of Na+ ions out of the cell and K+ ions into the cell. These actions occurring against their potential(concentration) gradients, which may be produced by hydrolyzing one ATP molecule.

Answer:

1/2 the distance

Explanation:

If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.

Answer:

No.Kerosene oil and water do not mix with each other and form two separate layers.

Answer:

No

Explanation:

They cannot be mixed together they will form upper and lower layer

Answer:

Time is 2.2 seconds.

Explanation:

Time:

[tex]{ \boxed{ \bf{time = \frac{distance}{speed} }}}[/tex]

Substitute into the formula:

speed = 715 km/h = 198.61 m/s

[tex]{ \tt{time = \frac{435}{198.61} }} \\ { \tt{time = 2.2 \: seconds}}[/tex]

Answer:

The amount of CH3OH present in the mixture would decrease

Explanation:

According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.

In this case, looking at the equation of the reaction:

2H2 + CO <------> CH3OH + heat

the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.

If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.

C₁₀H₁₄N₂ is the empirical formula of the compound whose mass is 40.57 present in 0.2500 moles.

Alkaloid compounds are those naturally present organic compounds in which nitrogen is present.

First we calculate the molecular mass of wanted compound by using the below formula:

n = W/M, where

n = no. of moles = 0.2500 moles (given)

W = given mass = 40.57 grams (given)

M = molar mass = 40.57 grams/= 0.2500 moles = 162.28 g/mole

In the question, given that:

% composition of carbon = 74.02%

% composition of hydrogen = 8.710%

% composition of nitrogen = 17.27%

Now we calculate the mass of these composition by using the below formula:

Composition mass = compound mass * % composition / 100

Mass of carbon = 162.28 * 74.02 / 100 = 120.11g

Mass of hydrogen = 162.28 * 8.710 / 100 = 14.13g

Mass of nitrogen = 162.28 * 17.27 / 100 = 28.02g

Now we calculate the moles of these composition to made empirical formula as:

Moles of carbon = 120.11 / 12 = 10

Moles of hydrogen = 14.13 / 1 = 14

Moles of nitrogen = 28.02 / 14 = 2

Thus, the empirical formula of the compound is C₁₀H₁₄N₂.

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After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

First, we will write the balanced equation for the reaction

H₂SO₄ + BaCl₂  → BaSO₄ + 2HCl

This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl

From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.

For H₂SO₄

mass = 50.0g

Molar mass = 98.079 g/mol

From the formula

Number of moles = Mass / Molar mass

∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol

Number of moles of H₂SO₄ = 0.5098 mol

For BaCl₂

mass = 40.0 g

Molar mass = 208.23 g/mol

∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol

Number of moles of BaCl₂ = 0.1921 mol

Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)

From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂

∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.

Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.

Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles

Now, we will convert this to grams

From the formula

Mass = Number of moles × Molar mass

Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol

Mass of H₂SO₄ that will remain = 31.1597 g

Mass of H₂SO₄ that will remain ≅ 31.16 g

Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

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Answer:

[tex]T_2= 36.7 \textdegree C[/tex]

Explanation:

Mass of Water [tex]m_w=6.90kg[/tex]

Temperature [tex]T=34.7 degrees[/tex]

Heat Flow [tex]H=57.1kJ[/tex]

Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]

Generally the equation for Final Temperature is mathematically given by

[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]

[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]

Therefore

[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]

[tex]T_2= 36.7 \textdegree C[/tex]

Answer:

Because it is dissolving and has pigment.  

Explanation:

Answer:

The solvated electron is responsible for a great deal of radiation chemistry. Alkali metals dissolve in liquid ammonia giving deep blue solutions which conduct electricity . The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.

Answer:

Electro positivity increases down the group

Answer:

[tex]n_1=4[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]

nf level= Balmer series

nf level= 2

Generally the equation for Wavelength is mathematically given by

[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]

Where

[tex]R=Rydberg Constant[/tex]

[tex]R=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]

[tex]n_1=4.0021[/tex]

[tex]n_1=4[/tex]

Answer:

Generation in computer terminology is a change in technology a computer is/was being used. Initially, the generation term was used to distinguish between varying hardware technologies. Nowadays, generation includes both hardware and software, which together make up an entire computer system.

5 generation of computer  

Main electronic component: based on artificial intelligence, uses the Ultra Large-Scale Integration (ULSI) technology and parallel processing method.

ULSI – millions of transistors on a single microchip

Parallel processing method – use two or more microprocessors to run tasks simultaneously.

Language – understand natural language (human language).

Power – consume less power and generate less heat.

Speed – remarkable improvement of speed, accuracy and reliability (in comparison with the fourth generation computers).

Size – portable and small in size, and have a huge storage capacity.

Input / output device – keyboard, monitor, mouse, trackpad (or touchpad), touchscreen, pen, speech input (recognise voice / speech), light scanner, printer, etc.

Example – desktops, laptops, tablets, smartphones, etc.

3 generation of computer

Main electronic component – integrated circuits (ICs)

Memory – large magnetic core, magnetic tape / disk

Programming language – high level language (FORTRAN, BASIC, Pascal, COBOL, C, etc.)

Size – smaller, cheaper, and more efficient than second generation computers (they were called minicomputers).

Speed – improvement of speed and reliability (in comparison with the second generation computers).

Input / output devices – magnetic tape, keyboard, monitor, printer, etc.

Examples – IBM 360, IBM 370, PDP-11, UNIVAC 1108, etc.

4 generations of computer

Main electronic component – very large-scale integration (VLSI) and microprocessor.

VLSI– thousands of transistors on a single microchip.

Memory – semiconductor memory (such as RAM, ROM, etc.)

RAM (random-access memory) – a type of data storage (memory element) used in computers that temporary stores of programs and data (volatile: its contents are lost when the computer is turned off).

ROM (read-only memory) – a type of data storage used in computers that permanently stores data and programs (non-volatile: its contents are retained even when the computer is turned off).

Programming language – high level language (Python, C#, Java, JavaScript, Rust, Kotlin, etc.).

Answer:

Help you with what hmm I do not know what you are talking about

Answer:

requiring immediate action or attention.

word similar to urgent: Acute

Answer:

The rate constant of an SN1 reaction depends on the nucleophile

The rate constant of an SN2 reaction does not depend on the nucleophile

Explanation:

Let us recall that in an SN1 reaction, the rate determining step involves only the alkyl halide substrate and not the nucleophile. Hence;

Rate = k[RX]

Therefore;

k= Rate/[RX]

For an SN2 reaction, the rate determining step involves both the nucleophile and the alkyl halide substrate.

Hence;

Rate = k[Nu-] [RX]

k= Rate/[Nu-] [RX]

Note that;

[Nu-] = concentration of the nucleophile

[RX] =concentration of alkyl halide substrate

k= rate constant

We can see from the above derivations that;

1) The rate constant of an SN1 reaction does not depend on the nucleophile

2) The rate constant of an SN2 reaction depends on the nucleophile

A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?

The pKa of acetic acid is 4.76.

Chemistry Buffer Calculations

1 Answer

Stefan V.

May 8, 2016

Δ

pH

=

0.29

Explanation:

!! LONG ANSWER !!

The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.

Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.

So, the Henderson-Hasselbalch equation looks like this

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

pH

=

p

K

a

+

log

(

[

conjugate base

]

[

weak acid

]

)

a

a

∣

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you have acetic acid,

CH

3

COOH

, as the weak acid and the acetate anion,

CH

3

COO

−

, as its conjugate base. The

p

K

a

of the acid is said to be equal to

4.76

, which means that you have

pH

=

4.76

+

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

The pH is equal to

5

, and so

5.00

=

4.76

+

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

=

0.24

This will be equivalent to

10

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

=

10

0.24

which will give you

[

CH

3

COO

−

]

[

CH

3

COOH

]

=

1.74

This means that your buffer contains

1.74

times more conjugate base than weak acid

[

CH

3

COO

−

]

=

1.74

×

[

CH

3

COOH

]

Now, because both chemical species share the same volume,

120 mL

, this can be rewritten as

n

C

H

3

C

O

O

−

120

⋅

10

−

3

L

=

1.74

×

n

C

H

3

C

O

O

H

120

⋅

10

−

3

L

which is

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

n

C

H

3

C

O

O

−

=

1.74

×

n

C

H

3

C

O

O

H

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

(

1

)

So, the buffer contains

1.74

times more moles of acetate anions that of acetic acid.

Now, the total molarity of the buffer is said to be equal to

0.1 M

. You thus have

[

CH

3

COOH

]

+

[

CH

3

COO

−

]

=

0.10 M

Once again, use the volume of the buffer to write

n

C

H

3

C

O

O

H

120

⋅

10

−

3

L

+

n

C

H

3

C

O

O

−

120

⋅

10

−

3

L

=

0.1

moles

L

This will be equivalent to

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

n

C

H

3

C

O

O

−

+

n

C

H

3

C

O

O

H

=

0.012

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

(

2

)

Use equations

(

1

)

and

(

2

)

to find how many moles of acetate ions you have in the buffer

1.74

⋅

n

C

H

3

C

O

O

H

+

n

C

H

3

C

O

O

H

=

0.012

n

C

H

3

C

O

O

H

=

0.012

1.74

+

1

=

0.004380 moles CH

3

COOH

This means that you have

n

C

H

3

C

O

O

−

=

1.74

⋅

0.004380 moles

n

C

H

3

C

O

O

−

=

0.007621 moles CH

3

COO

−

Now, hydrochloric acid,

HCl

, will react with the acetate anions to form acetic acid and chloride anions,

Cl

−

H

Cl

(

a

q

)

+

CH

3

COO

−

(

a

q

)

→

CH

3

COO

H

(

a

q

)

+

Cl

−

(

a

q

)

Notice that the reaction consumes hydrochloric acid and acetate ions in a

1

:

1

mole ratio, and produces acetic acid in a

1

:

1

mole ratio.

Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

c

=

n

solute

V

solution

⇒

n

solute

=

c

⋅

V

solution

a

a

∣

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, this gets you

n

H

C

l

=

0.300 mol

L

−

1

⋅

volume in liters

6.60

⋅

10

−

3

L

n

H

C

l

=

0.001980 moles HCl

The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain

n

H

C

l

=

0 moles

→

completely consumed

n

C

H

3

C

O

O

−

=

0.007621 moles

−

0.001980 moles

=

0.005641 moles CH

3

COO

−

n

C

H

3

C

O

O

H

=

0.004380 moles

+

0.001980 moles

=

0.006360 moles CH

3

COOH

The total volume of the solution will now be

V

total

=

120 mL

+

6.60 mL

=

126.6 mL

The concentrations of acetic acid and acetate ions will be

[

CH

3

COOH

]

=

0.006360 moles

126.6

⋅

10

−

3

L

=

0.05024 M

[

CH

3

COO

−

]

=

0.005641 moles

126.6

⋅

10

−

3

L

=

0.04456 M

Use the Henderson-Hasselbalch equation to find the new pH of the solution

pH

=

4.76

+

log

(

0.04456

M

0.05024

M

)

pH

=

4.71

Therefore, the pH of the solution decreased by

Δ

pH

=

|

4.71

−

5.00

|

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

0.29 units

a

a

∣

∣

−−−−−−−−−−−−−

Answer link

Related topic

Buffer Calculations

Questions

Related questions

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How do you buffer a solution with a pH of 12?

Why are buffer solutions used to calibrate pH?

What is the role of buffer solution in complexometric titrations?

How you would make 100.0 ml of a 1.00 mol/L buffer solution with a pH of 10.80 to be made using...

What is the Henderson-Hasselbalch equation?

What is an example of a pH buffer calculation problem?

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Answer:

Solution temperature.

Explanation:

Hello there!

In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:

1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.

2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.

In such way, we infer the answer is solution temperature.

Regards!

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. Moles ethanol -Molar mass: 46.07g/mol-:

25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol

kg solvent:

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

Molality:

0.54265 moles ethanol / 0.075kg = 7.24m

b. Liters solution:

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L

Molarity:

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

Answer:

с

Explanation:

the first quantum number of an electron gives the information about the energy level the electron is in

The information first quantum number of an electron give is the energy level the electron is in.

Quantum numbers is a set of symbols which gives idea about the position  of electron present inside an atom.

Hence first quantum number of electron gives ideal about the energy level.

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Answer:

Volume = 1222.5cm³

Explanation:

If the question is about the volume of the rectangle:

The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.

In the problem, the length of the rectangle is 0.162m = 16.2cm

The width is 7.7cm

And the height is 9.8cm

The volume is:

Volume = 16.2cm*7.7cm*9.8cm

Answer:

In Haloarenes the C atom to which the X group is attached is SP2 hybridized thus it is become difficult to replace it by the Nucleophile. Since arenes and Vinyl halides are electron rich molecules due to presenceof n bonds, they repel Nucleophile attacking them.

Question:-

Why it is difficult for haloarenes to undergo nucleoplhilic subsituⁿ reaction?

Answer:-

Haloarenes are less reactive towards the nucleoplhilic substitution rxⁿ . This is due to following reasons :-

[tex]\red{\bigstar}\underline{\textsf{ Reason 1 :- Partial double bond character .}}[/tex]

Halogen atom has one lone pair, and due to presence of π - σ - lp , resonance is established in the compound ( see attachment) . Due to resonance there is a partial double bond character in the carbon halogen bond , so it is difficult to break a double bond than a single bond.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 2 :- $\pi$ cloud .}}[/tex]

When a nucleoplhile comes to attack , it is repelled by the π-cloud of the benzene ring.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 3 :- Phenyl cation .}}[/tex]

If somehow the halogen atoms leaves the benzene ring ,being more electronegative than carbon , it takes away the electron , thus a positive charge is left on benzene ring and the phenyl cation so formed is very unstable .

[tex]\rule{200}2[/tex]

Answer:

To prevent the coagulation of blood

Explanation:

Sodium chloride oxalate in blood sample containers exists to prevent the coagulation of blood samples being collected (for analytical purposes).

The compound works by chelating or combining with blood's calcium which is necessary for blood to coagulate (change from flowing liquid to gel-like substance).

Calcium ions are known to be responsible for the activation of certain coagulation factors such as FXIII and therefore, play important roles in homeostasis during blood clotting. Hence, the removal of the calcium in blood disrupts the homeostatic process of clotting formation and results in the blood not being able to coagulate.

Answer: Promotion of electrons is accompanied by a release of energy because of absorption of photon.

Explanation:

Promotion of electrons occurs when an electron accepts or absorbs a photon which leads to it's movement from a lower energy level orbital to a higher energy orbital.

According to Bohr, the electrons was restricted to certain energy levels and was thought to move along certain circular orbits around the nucleus. These energy levels were identified by means of principal quantum number, n. The wave mechanics model of atom does not restrict the electrons to a certain energy levels only. Instead it describes a region around the Nucleus called orbitals where there is a possibility of finding an electron with a certain amount of ENERGY.

The energy levels are composed of one or more orbitals and the distribution of electrons around the nucleus is determined by the number and kind of energy levels that are occupied.

Bohr made an assumption that an electron emits energy in the form of radiation when it moves from a higher to a lower permitted orbit, this produces a line in the atomic emission spectrum. Since the energies of the higher and lower orbits are fixed, the line will be of a particular energy and frequency.

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

Learn more about dilution, here:

https://brainly.com/question/15718488

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Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

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