Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the C-side of L-Arg and L-Lys, but not on the C-side of D-Arg and D-Lys. Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the -side of L- and L-, but not on the -side of D- and D-. All the naturally occurring amino acids are D-amino acids, while active centres of serine proteases are nearly all L. All the naturally occurring amino acids are L-amino acids, while active centres of serine proteases are nearly all D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site. None of the above.

Answer:

the number of milliliters of a 1M is 402mL

Explanation:

The computation of the number of milliliters could be determined by using the following formula

As we know that

[tex]V_1\times M_1 = V_2\times M_2[/tex]

where,

V_1 and V_2 are the starting and final volumes

And, the M_1 and M_2 are the starting and the final molarities

Now the V_1 is

[tex]V_1 \times 1M = 60mL \times 6.7M[/tex]

So, the V_1 is 402mL

Hence, the number of milliliters of a 1M is 402mL

Answer:

Heterogeneous Mixture. Have a good day! =)

Explanation:

Answer:

the answer is C.

Explanation:

this is because burning anything is going to change the make-up of the object

Answer:

physical projects

Journal entries

Materials

Lab reports

Artworks

Explanation:

Definition of a Portfolio:

Portfolio can be defined as a physical collection of student work that includes materials such as written assignments, completed tests, artwork, lab reports, physical projects and other material evidence of learning progress and academic accomplishment, including awards and honors,

A portfolio is a long-term form of self reflection and assessment that students do together.

Portfolios are a great way to demonstrate the competencies you would list on a resume or talk about in a science interview

Answer:

FLUORINE, CHLORINE, BROMINE, IODINE, ASTATILE AND TENNESINE

Explanation :

Florine and chorine=gas

Bromine=liquid

ASTATILE and TENNESINE =solid

HOPE IT is right

Answer:

The formation of an aqueous solution of ammonium nitrate is An endothermic process

Explanation:

An exothermic process produce energy when occurs. As there is energy that is released, the temperature of the arounds increases.

In the other hand, an endothermic process absorb energy when occurs doing the temperature of the around colder than the initial temperature.

As the dissolution of NH₄NO₃ in water make the temperature of the water colder:

The formation of an aqueous solution of ammonium nitrate is an endothermic process

The question requires us to determine if reaction process is an endothermic or an exothermic reaction.

To do this,

First we will define the terms Endothermic reaction and exothermic reaction

Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. They have a net positive standard enthalpy change.

Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light. They have a net negative standard enthalpy change.

From the question,

When solid NH4NO3 is dissolved in water, the temperature of the water and beaker gets noticeably colder. This means it is an endothermic reaction.

Hence, the formation of an aqueous solution of ammonium nitrate is an endothermic process

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Answer:

[tex]\Huge \boxed{\mathrm{C_2H_5OH}}[/tex]

Explanation:

CH₃OCH₃ (Dimethyl ether) has a boiling point of -23 °C

C₂H₅OH (Ethanol) has a boiling point of 78.37 °C

CH₃CH₂CH₃ (Propane) has a boiling point of -42 °C

CH₃CHO (Acetaldehyde) has a boiling point of 20.2 °C

C₂H₅OH (Ethanol) has the highest boiling point.

Answer:

the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Explanation:

To calculate the solubility of BaCO₃ in:

(a) pure water and (b) in a solution in which [CO₃²⁻] = 0.289 M

The [tex]ksp[/tex] (i.e solubility constant ) for BaCO₃= 2.0 × 10⁻⁹

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

s² = ksp

s = [tex]\sqrt{ksp}[/tex]

s = [tex]\sqrt{2.0 \times 10^{-9}}[/tex]

s = 4.472 × 10⁻⁵ M

(b) The solubility of BaCO₃ in a solution in which [CO₃²⁻] = 0.289 M

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

2.0 × 10⁻⁹ = s × 0.289

s = 2.0 × 10⁻⁹/0.289

s = 6.9204 × 10⁻⁹ M

Thus, the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Answer:

Molality of the methanol solution = 8.33 m

Explanation:

Given:

Mass % = 21.1 %

Molar mass of methanol = 32.04 g / mol

Find:

Molality of the methanol solution?

Computation:

Moles of methanol = Mass / Molar mass

Moles of methanol =  = 21.1 / 32.04

Moles of methanol = 0.658

Assume.

Mass of solution = 100 g

Mass of solvent  = 100 -21.1 = 78.9 g  = 0.0789 kg

Molality of the methanol solution = 0.658 / 0.0789

Molality of the methanol solution = 8.33 m

Given values:

Now,

Moles of methanol:

= [tex]\frac{Mass}{Molar \ mass}[/tex]

= [tex]\frac{21.1}{31.04}[/tex]

= [tex]0.658[/tex]

then,

Mass of solution:

= [tex]100-21.1[/tex]

= [tex]78.9 \ g \ or \ 0.0789 \ kg[/tex]

hence,

The molality will be:

= [tex]\frac{0.658}{0.0789}[/tex]

= [tex]8.33 \ m[/tex]

Thus the answer above is correct.

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Answer:

Because the model makes the concepts easier to understand

Explanation:

Models are created to give a visual of every aspect of an experiment. This allows for a better understanding across the board for everyone.

Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

Answer:

pH of the final solution = 9.15

Explanation:

Equation of the reaction: HCl + NH₃ ----> NH₄Cl

Number of moles of  NH₃ = molarity * volume (L)

= 0.4 M * (300/1000) * 1 L =  0.12 moles

Number of moles of HCl =  molarity * volume (L)

= 0.3 M * (175/1000) * 1 L = 0.0525 moles

Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced

Number moles of NH₄Cl produced =  0.0525 moles

Number of moles of base left unreacted =  0.12 - 0.0525 = 0.0675

pOH = pKb + log([salt]/[base])

pKb = -logKb

pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)

pOh = 4.744 + 0.109

pOH = 4.853

pH = 14 - pOH

pH = 14 - 4.853

pH = 9.15

Therefore, pH of the final solution = 9.15

Answer:

i think its natural . it is the natural process

Answer:

Solar radiation falls into the category of natural forces.

Explanation:

Natural forces don’t have a strong effect on recent global temperatures. The natural forces line begins to drop around 1950 and does not track with the observed temperature line. On the other hand, the human effects line rises upward with the observed temperatures, so it’s a stronger effect.

Brainilest?

The question is incomplete, so the complete question is as follows:

Decide which element probably has a melting point most and least similar to the melting point of cesium. Comparing melting point: helium thallium sodium strontium

Answer:

Most similar melting point : Sodium

Least similar melting point: Strontium

Explanation:

Melting point is defined as the temperature at which soid will melt or get converted into liquid.

Melting point of cesium is 28.44 °C and most similar melting point  is sodium as its melting point is 97.72°C and least similar melting point is strontium as it has melting point 777°C.

Hence, the correct answer is:

Most similar melting point : Sodium

Least similar melting point: Strontium

Answer:

The change in temperature is [tex]26.06^{\circ} C[/tex].

Explanation:

It is given that,

The specific heat of Aluminium is cal/g°C

Mass of the sample, m = 4.55 g

Heat absorbed, Q = 2.55 cal

We need to find the change in temperature of the sample. The heat absorbed by an object is given by :

[tex]Q=mc\Delta T[/tex]

[tex]\Delta T[/tex] is the change in temperature

So,

[tex]\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{2.55\ cal}{4.55\ g\times 0.0215 \ cal/g^{\circ} C}\\\\\Delta T=26.06^{\circ} C[/tex]

So, the change in temperature is [tex]26.06^{\circ} C[/tex].

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

[tex]heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)[/tex]

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

Answer:

[tex]h=1.60\mu m[/tex]

Explanation:

Hello,

In this case, given the density and the mass of the aluminum foil, we can compute the occupied volume as shown below:

[tex]\rho =\frac{m}{V}\\ \\V=\frac{m}{\rho}=\frac{2.07g}{2.7g/cm^3} =0.767cm^3[/tex]

Next, since the volume is defined as:

[tex]V=24cm*20cm*h[/tex]

Whereas [tex]h[/tex] accounts for its thickness, we can find it to be:

[tex]h=\frac{V}{24cm*20cm}=\frac{0.767cm^3}{20cm*24cm}\\ \\h=1.60x10^{-3}cm*\frac{10000\mu m}{1cm} \\\\h=1.60\mu m[/tex]

Regards.

Answer:

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)

Explanation:

When we want to balance redox reaction equations, we must ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation.

After we have done this, we can now write the overall balanced reaction equation without including the number of electrons lost or gained. Hence;

Oxidation half equation;

3Sb^3+(aq) -----> 3Sb^5+(aq) +6e

Reduction half equation;

BrO3^-(aq) + 6H^+(aq) + 6e ----> Br^-(aq) 3H2O(l)

Overall balanced reaction equation;

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)

Answer:

b. In equilibrium with the liquid phase, the pressure exerted by a gas.

Explanation:

When a liquid is warmed up to a temperature , it starts vaporising . The liquid is turning into gas and gas is turning into liquid at different rates  . Initially the rate of former is higher but gradually the difference of rate between them decreases to zero . At this point the rate of conversion of liquid into gas and rate of conversion of gas into liquid becomes equal  . This is called dynamic equilibrium point .

If we change the temperature , the equilibrium gets disturbed .

At this point the pressure exerted by the gas is called the vapour pressure of the liquid .

So option b ) is correct .

At the equilibrium point, the pressure exerted by the gas molecules to the liquid molecules has been termed the Vapor pressure. Thus, statement b is correct.

The vaporization has been the process of conversion of liquid to the gaseous state with the rise in temperature. The liquids attaining a certain temperature have been vaporized into the gaseous state.

Initially, the gas phase has been less in concentration, thus the rate of formation of gas has been greater.

After a certain amount of time, the gas phase starts to cool down and converts to the liquid state. The rate of formation of the liquid has been slower.

The time when the rate of formation of liquid, and the rate of formation of gas has been equal is termed as the equilibrium point. At the equilibrium point, the pressure exerted by the gas molecules to the liquid molecules has been termed the Vapor pressure.

Thus, statement b is correct.

For more information about the Vapor pressure, refer to the link:

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Answer:

C). The previously known outcome of an observation or experiment can be used as solid proof that a newly-created scientific hypothesis is absolutely true.

Explanation:

A Scientific hypothesis is characterized as the proposed explanation or an educated guess about a natural phenomenon on the basis of previous knowledge as well as observation. All the given statements are true regarding a scientific hypothesis that it is 'an educated guess which explains the reason why a specific phenomenon occurs', 'being testable in a manner that could be replicated by other', 'an explanation for a natural phenomenon' except for that  the 'truth of a scientific hypothesis can never be assured completely with a solid proof as it always has chances of being expanded.' Thus, option C is the correct answer.

Answer:

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Explanation:

In first place, the balanced reaction between Mg and O₂ is:

2 Mg + O₂ ⇒ 2 MgO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

[tex]moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}[/tex]

moles of MgO= 0.250

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Answer:

D) 5

Explanation:

In this case, the "monobromination" is the addition of one "Br" to the molecule. With this in mind, we have to explore each option to put this "Br" atom:

1) (bromomethyl)cyclopentane = The Br atom is placed in carbon 6

2) 1-bromo-1-methylcyclopentane = The Br atom is placed in carbon 1

3) (1R,2R)-1-bromo-2-methylcyclopentane = The Br atom is placed in carbon 2 (with trans configuration)

4) (1R,2S)-1-bromo-2-methylcyclopentane = The Br atom is placed in carbon 2 (with cis configuration)

5) 1-bromo-3-methylcyclopentane = The Br atom is placed in carbon 3

See figure 1

I hope it helps!

Answer:

Yes, given statement is true.

Explanation:

Given that a cylinder of 1000 ml is marked at every 100 ml.

During the test, the student can easily check the volume of liquid so the student's guess is 750 ml true.  

With this measurement, we can estimate that 1/10th of 100 ml can be measured.  

So, given statement is correct.

Answer:

Yes, the digit in the hundreds place is certain and the digit in the tens place is estimated.

Explanation:

Answer:

Explanation:

Density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

From the question

mass = 145.8 g

volume = 91.75 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{145.8}{91.75} [/tex]

= 1.5891

We have the final answer as

Density = 1.6 g/mL

Hope this helps you

Answer:

pure substances and mixtures

Explanation:

pure substances are further broken down into elements and compound....... A mixture is composed of different types of atoms or molecules that are not chemically bonded.

Heterogeneous and homogeneous mixtures are the two types of mixtures. While homogeneous mixtures seem consistent throughout, heterogeneous mixtures have clearly discernible components. A solution, which can be a solid, liquid, or gas, is the most typical kind of homogenous mixture.

Mixtures are materials made up of two or more different types of matter. Physical means can be used to separate them. Examples include a salt-water solution, a sugar-water solution, various gases, air, etc. The many components of any mixture do not come together by any sort of chemical transformation.

Solutions, suspensions, and colloids are the three categories into which mixtures can be divided based on particle size. A mixture's constituent parts maintain their unique physical characteristics

Answer:

both pure materials and mixes

Explanation:

The breakdown of pure substances into their component components and compounds continues. Different kinds of unchemically linked atoms or molecules make up a mixture.

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Answer:

[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]

Explanation:

Using the approach of Henderson-HasselBalch equation, we have :

[tex]pH = pKa[NH^+_4] + log \dfrac{[NH_3]}{[NH_4^+]}[/tex]

where;

the pKa of [tex]NH^+_4[/tex] = 9.26

concentration of [tex]NH_3[/tex] = 0.100 M

concentration of [tex]NH_4Cl[/tex] = 0.176 M

∴

the pH of the buffered solution is :

[tex]pH = 9.26 + log \dfrac{[0.100]}{[0.176]}[/tex]

[tex]pH = 9.26 + log (0.5682)[/tex]

[tex]pH = 9.26 +(-0.2455)[/tex]

[tex]pH =9.02[/tex]

The Chemical equation for the reaction of [tex]Zn ^{2+}[/tex] and EDTA is :

[tex]Zn^{2+}_{(aq)} + Y^{4-}_{(aq)} \iff ZnY^{2-} _{(aq)}[/tex]

Here;

[tex]Y^{4-}_{(aq)}[/tex] denotes the fully deprotonated form of the EDTA

The formation constant [tex]K_f[/tex] of the equation for the reaction can be represented as:

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex]      ----- (1)

The logarithm of the formation constant of Zn - EDTA complex = 16.5

[tex]K_f[/tex]  = [tex]10^{16.5}[/tex]

[tex]K_f[/tex]  = [tex]3.16 \times 10^{16}[/tex]

Since the formation constant in the above equation signifies that the EDTA is present in  [tex]Y^{4-}[/tex],

Then:

[tex]\alpha _{Y^{4-} }= \dfrac{Y^{4-}}{C_{EDTA}}[/tex]

[tex]{Y^{4-}}= \alpha_ {Y^{4-}} \times {C_{EDTA}}[/tex]

From (1)

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex]  

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ \ \alpha_ {Y^{4-}} \times {C_{EDTA}}}[/tex]

∴

[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

where;

[tex]K_f'[/tex] = conditional formation constant

[tex]\alpha _Y{^4-}[/tex] = the fraction of EDTA that exit in the form of the presences of the 4 charges .

So at equivalence point :

all the [tex]Zn^{2+}[/tex] initially in titrand  is now present in [tex]ZnY^{2-}[/tex]

[tex]K_f' = K_f \times \alpha _Y{^4-}[/tex]

Obtaining the data for the value of [tex]\alpha _Y{^4-}[/tex] at the reference table:

[tex]\alpha _Y{^4-}[/tex]  =  [tex]5.4 \times 10^{-12}[/tex]

∴

[tex]K_f' = 3.16 \times 10^{16} \times 5.4 \times 10^{-2}[/tex]

[tex]K_f' = 1.7064 \times 10^{15}[/tex]

To calculate the moles of  EDTA ,[tex]Zn^{2+}[/tex]  , [tex]ZnY^{2-}[/tex] ; we have:

moles of  EDTA = 0.0100 M × 0.025 L

moles of  EDTA = [tex]2.5 \times 10^{-4} \ mole[/tex]

moles of [tex]Zn^{2+}[/tex] = 0.00500 M  × 0.050 L

moles of [tex]Zn^{2+}[/tex] = [tex]2.5 \times 10^{-4} \ mole[/tex]

moles of  [tex]ZnY^{2-}[/tex]  =  [tex]\dfrac{initial \ mole}{total \ volume}[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.025 + 0.050 }[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.075 }[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = 0.0033333 M

Recall that:

[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

[tex]K_f' = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

Assume Q² is the amount of complex dissociated in [tex]ZnY^{2-}[/tex]

[tex]ZnY^{2-} \iff Zn^{2+} + C_{EDTA}[/tex]  

i.e [tex]Q^2 = Zn^{2+} + C_{EDTA}[/tex]

[tex]1.707 \times 10^{15}= \dfrac{0.0033333}{Q}[/tex]

[tex]Q= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]

[tex]Q^2= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]

[tex]Q^2= 1.9527 \times 10^{-18}[/tex]

[tex]Q= \sqrt{1.9527 \times 10^{-18}}[/tex]

Q = [tex]1.397 \times 10^{-9}[/tex] M

[tex][Zn^{2+}]= 1.39 \times 10^{-9} \ M[/tex]

∴

[tex]pZn ^{2+} =- log [Zn^{2+}][/tex]

[tex]pZn ^{2+} = - log (1.39 \times 10^{-9} ) \ M[/tex]

[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]