Amphiphilic molecule: ___________

a. have both oxidizing and reducing groups.
b. are micelles.
c. have chromophores in two different wavelength regions.
d. have both acidic and basic groups.
e. have both hydrophilic and hydrophobic groups.

Answers

Answer 1
The answer is E - have both hydrophilic and hydrophobic groups.

Related Questions

equal volumes 50.0 ml of 0.54897 m solution of HCL and 0.9573 M NaOH solution have an initial temperature of 18.5 degrees Celsius react in a coffee cup calorimeter the resulted Solutions records a temperature of 22.9 degrees Celsius since the solutions are mostly water the solution are assumed to have a density of 1.0 g per ml and a specific heat of 4.184 joules per gram degree celsius the heat capacity of the Colorimeter is 1.0 x 10 to power 1 joules per degrees Celsius calculate the heat and gained or released by the solution at a constant pressure in J

Answers

Answer:

Explanation:

ok

Based on the crystal- field strength cl-

Answers

1) Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field igand and would cause minimum splitting of d orbitals.

The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.

Answers

Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.

The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

I hope it helps!

Given that H2(g)+F2(g)⟶2HF(g)ΔH∘rxn=−546.6 kJ 2H2(g)+O2(g)⟶2H2O(l)ΔH∘rxn=−571.6 kJ calculate the value of ΔH∘rxn for

Answers

Answer:

ΔH∘rxn of the reaction is -521.6kJ

Explanation:

Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"

You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:

(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ

(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ

Subtracting 2ₓ(1) - (2)

2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ

-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ

2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)

H₂(g) are cancelled because are the same in products and reactants

ΔH∘rxn = -1093.2kJ + 571.6kJ

ΔH∘rxn = -521.6

ΔH∘rxn of the reaction is -521.6kJ

Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?

Answers

cyclohexane will distill off first as it will have lower boiling point compared to ortho xylene which has higher molecular mass

pls help!!! How can you prove the pennies are made of different material aside from cutting them in half?

Answers

Answer:

Most of our coins are metal sandwiches. The outside layers are three-quarters copper and one-quarter nickel, and the "filling" is solid copper. Pennies are made of zinc coated with copper. Only nickels are one solid material—that same 75% copper/25% nickel alloy

Explanation:

Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?

Answers

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Match each term to an appropriate example. (4 points) Column A 1. Gravitational energy : Gravitational energy 2. Nuclear energy : Nuclear energy 3. Radiant energy : Radiant energy 4. Stored mechanical energy : Stored mechanical energy Column B a. Fission and fusion b. A ball at the top of a hill c. A compressed spring d. X-rays and light

Answers

Answer:

Explanation:

Before proceeding to answering the questions, let's try and understand each term.

Gravitational energy can be defined as the potential energy an object has when placed in a high position (compared to been placed in a low position). Examples include; a knife been placed at the top of the cupboard, a ball at the top of a hill.

Nuclear energy can be defined as the energy released or consumed as a result of splitting or joining together of the nuclei (plural of nucleus) of atom(s). The process of splitting an atomic nucleus is called nuclear fission while the process of joining two nuclei of atoms is called nuclear fusion.

Radiant energy can be defined as the energy of electromagnetic wave. One of the properties of electromagnetic waves is that they travel through space. Examples include X-rays and light

Stored mechanical energy can be defined as the potential energy an object has as a result of the application of force. Examples include; a stretched rubber band, a compressed spring

03/08/2020
Question
1. (a) State the definition of a chemical formula.
(b) What does it tell about a compound?
(c) What information is conveyed by the formulation H2SO4?​

Answers

Explanation:

According to your question,

no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.

b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.

c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).

Hope it helps...

What are the conjugate acid-base pairs in the following chemical reaction? HBr(aq)+ CH3COOH(aq) ⇌ CH3C(OH)2+(aq) + Br-(aq)

Answers

Answer:

HBr, CH3C(OH)2 and CH3COOH, Br-

Explanation:

The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.

Conjugate acid is formed when a bases receives a proton  (H+) and a conjugate base is formed when an acid losses a proton (H+).

From the given equation:

HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.

Gas is contained in a 9.00-L vessel at a temperature of 24.0°C and a pressure of 5.00 atm. (a) Determine the number of moles of gas in the vessel. mol (b) How many molecules are in the vessel? molecules

Answers

Answer:

a. Moles in the vessel = 1.85 moles of the gas

b. 1.11x10²⁴ molecules are in the vessel

Explanation:

a.It is possible to determine moles of a gas using the general law of gases:

PV = nRT

Where P is pressure: 5.00atm; V is volume = 9.00L; R is gas constant: 0.082atmL/molK; T is absolute temperature: 273.15K +24.0 = 297.15K

Computing the values:

PV / RT = n

5.00atm* 9.00L / 0.082atmL/molK*297.15K = n

Moles in the vessel = 1.85 moles of the gas

b. With Avogadro's number we can convert moles of any compound to molecules thus:

Avogadro's number = 6.022x10²³ molecules / mole

1.85moles ₓ (6.022x10²³ molecules / mole) =

1.11x10²⁴ molecules are in the vessel

15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass

Answers

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

Cho biết độ tan của NH4Cl trong nước ở 20oC và 70oC lần lượt là 37,2 g/100 gam nước và 60,2 gam/100 g nước. Hòa tan 166,8 gam NH4Cl vào 400 gam nước ở 70oC thu được dung dịch X. Sau đó, hạ nhiệt độ dung dịch X xuống 20oC. Tính khối lượng (gam) NH4Cl kết tinh lại trong X?

Answers

Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C

Actini(III) hydroxide Ac(OH)3   0,0022    

Amonia NH3 1176 900 702 565 428 333 188

Amoni azua NH4N3 16  25,3  37,1  

View 42 more rows  

                    hehe

In which ONE of the following compounds would the bonding be expected to have the highest percentage of ionic character? A) LiBr B) CsCl C) BaBr2 D) NaCl E) KI

Answers

Answer:

B) CsCl

Explanation:

The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.

In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).

The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.

Which two layers are part of the thermosphere?

O exosphere and ionosphere

O ionosphere and mesosphere

mesosphere and stratosphere

O stratosphere and troposphere

Answers

The two layers are part of the thermosphere are  exosphere and ionosphere.

What is the exosphere layer?

The Exosphere is the topmost layer of the Earth's atmosphere.

and its gradually disappear into the vacuum of space.

It consist two parts that are:

exosphere and ionosphere.

Thus, option "A" is correct, the rest of the option is not a part of thermosphere.

To learn more about  atmospheric layers click here:

https://brainly.com/question/25881294

#SPJ2

Answer:

hi hope your doing great the answer is A

Explanation:

its on Edge 2020

hope i helped :)

What is the mass grams that are in 3.52 × 10²⁵ molecules of I₂

Answers

Answer:

As you know, one mole of any substance contains exactly

6.022

10

23

molecules of that substance - this is known as Avogadro's number.

Notice that you're dealing with more than

6.022

10

23

molecules of carbon dioxide, which means that you'll also be dealing with more than one mole of the compound.

More specifically, you'll have

1.5

10

26

molecules

1 mole CO

2

6.022

10

23

molecules

=

2.491

10

2

moles CO

2

Now, a substance's molar mass tells what the mass of one mole of that substance is. In carbon dioxide's case, its molar mass is equal to

44.01 g/mol

, which means that every mole of

CO

2

will have a mass of

44.01 g

.

In your case,

2.491

10

2

moles of

CO

2

would have a mass of

2.491

10

2

moles CO

2

44.01 g

1

mole CO

2

=

109.63 g

Rounded to two sig figs, the number of sig figs you have for the number of molecules of

CO

2

, the answer will be

m

C

O

2

=

110 g

Compare the modern (electron cloud) model of the atom with Bohr’s atomic model. Which of these statements describe the two models correctly? Check all of the boxes that apply.

A. Bohr’s model was replaced only because of its age.

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model the location of the electrons cannot be predicted.

C. The modern model explains all available data about atoms; Bohr’s model does not.

D. The modern model is more widely accepted because it was proposed by more well known scientists.

Answers

Answer:

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model, the location of the electrons cannot be predicted.

AND

C. The modern model explains all available data about atoms; Bohr’s model does not.

Explanation:

The answers are right on Edge. :)

Answer:

b and c

Explanation:

my assignment was 100% 2022

Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction

Answers

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing two similarities and two differences.

Answers

Answer:

hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

answer : It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

They have different Geometry

They differ in bond angles as well

Explanation:

The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

comparing(similarities) these two generic molecules

It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

differences between the generic molecules

They have different Geometry

They differ in bond angles as well

Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)

Answers

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  


1. What are some examples of forces found in everyday life? (Give examples from the bike picture
above OR make up your own examples.)
2 Define the word "motion"...
3. What do you think the following phrase from the definition means with respect to time and in
comparison to the position of other objects used as reference points?"
4. Define the word "Yorco...
5. Give an example from everyday life of an object that is "in motion (there is no "correct answer for
this - all reasonable answers accepted).
6. What is an example from everyday life of a force acting on an object? (there is no correct answer"..)
i need help on the questions

Answers

Answer:

1.answer = pushing force

= pulling force

=gravity force

Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?

Answers

Yes beautiful language

Which of the following chemical equations corresponds to the standard molar enthalpy of formation of Na_2CO_3(s)?
a. 2 NA(s) + C(s) + 3 O(g) ------------> Na_2CO_3(s)
b. Na_2O(s) + CO_2(g) --------------->Na_2CO_3 (s)
c. Na_2(s) + C(s) + 3 O(g) -------------> Na_2CO_3 (s)
d. Na_2O(s) + CO(g) ---------------> Na_2CO_3(s)
e. 2 Na(s)+ C(s) + 3/2 O_2(g) ------------> Na_2CO_3(s)

Answers

Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Which is the best interpretation of the two flat portions of the graph?

Answers

In those portions, the heat supplied to the substance by the heater does not lead to a temperature rise as intermolecular forces are broken.

When a substance is heated, we normally expect that its temperature will rise as a consequence.

However, heat may be supplied to a substance but its temperature does not rise owing to the fact that the heat energy supplied is used to break intermolecular bonds.

This occurs during fusion and boiling. The heat supplied at these point does not result in temperature rise since it is used to break intermolecular bonds. The temperature remains steady during these processes as shown by the two flat portions on the graph in the image attached to the question. This heat supplied is known as the latent heat.

For more about latent heat, see:

https://brainly.com/question/19863536

Answer:

The answer for me was "These show where changes of state are occuring"

Explanation:

I got it right

Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN

Answers

Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

ABCDBCEFCDEF

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

H2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.

Answers

Answer: [tex]\Delta S[/tex] = 473.92J/K.mol

Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:

[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]

The balanced reaction:

[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]

gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:

3.2 moles of water is used;

4.8 moles of hydrogen gas is formed;

1.6 moles of sulfur dioxide is also formed;

Calculating entropy change:

[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]

[tex]\Delta S=628.8+398.08-328.96-224[/tex]

[tex]\Delta S[/tex] = 473.92J/K.mol

Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol

When the nuclide carbon-14 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is

Answers

Answer:

Explanation:

The equation for the decay is given as;

¹⁴₆C  --> X + ⁰₋₁e

For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;

Mass number;

14 = x + 0

x = 14

Atomic Number;

6 = x + (-1)

x = 6 + 1 =7

¹⁴₆C  --> ¹⁴₇N + ⁰₋₁e

The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N

If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.

Answers

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

2.5
Nakula investigated the effect of heat on amylase. Amylase is an enzyme
that makes starch molecules break down into sugar molecules.
P
• Nakula put some amylase solution into two boiling tubes,
P and Q
• He boiled the solution in tube P. He did not heat tube Q.
• He waited until the solution in tube P had cooled down to
room temperature.
• He added equal volumes of starch solution to tube P and
boiled amylase amylase
• After 10 minutes, he tested both tubes for sugar.
and starch and starch
• Nakula found that there was sugar in tube Q. but not in tube P.



Answers

The structure of amylase deteriorates due to high temperature of the solution.

This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.

At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.

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Nakula's conclusion was- "My results show that boiling destroys amylase"

Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.

Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.

Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significant digits.

Answers

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

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