What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase

Answer:

2

Explanation:

Lead(|V) fluoride

Ammonium Nitrate

Lithium sulfide

For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.

The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.

The second one is just two polyatomic ions which you just have to remember.

The last one is the typical ionic compound naming technique i guess.

Answer:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Explanation:

Hey there!

In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Regards!

Explanation:

the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet

Answer:

[tex][O_2]_{eq}=0.030M[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:

[tex][O_2]_{eq}=0.050M-x[/tex]

Whereas [tex]x[/tex] can be found considering the equilibrium of SO3:

[tex][SO_3]_{eq}=2x=0.040M[/tex]

Which means:

[tex]x=\frac{0.040M}{2} =0.020M[/tex]

Thus, the equilibrium concentration of oxygen gas turns out:

[tex][O_2]_{eq}=0.050M-0.020M=0.030M[/tex]

Regards!

Answer:

Fe(s)->Fe2+2e-

Explanation:

A.p.e.x

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

Oxidation reaction is a chemical reaction which can be described as follows ;

In the given reaction ;

Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.

Therefore,

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

Learn More about redox reaction here ;

https://brainly.com/question/13978139

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Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3

b) When we include states of matter;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)

c) Complete ionic equation;

2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)

Net Ionic equation;

Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)

The question is incomplete, the complete question is;

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon

a) electronic excitation

b) bond breakage

c) molecular vibration

d) molecular rotation

Answer:

molecular rotation

Explanation:

Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.

When molecules absorb infrared radiation, they transition between the rotational states of the molecule.

Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

D.22

is my answer than welcome

Answer:

5×6.02×1023

Explanation:

there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010

Answer:

[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

[tex]d=\frac{m}{V}[/tex]

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

[tex]m=552.4g-464.7g=87.7g[/tex]

So that we are now able to calculate the density in g/mL first:

[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]

Now, we proceed to the conversion to lb/in³ by using the following setup:

[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]

Regards!

Answer:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:

[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]

Now, we can separate the nitrates in ions as they are aqueous to obtain:

[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]

And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Best regards!

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

Answer:

1.27 × 10⁵ L

Explanation:

Step 1: Given data

Step 2: Convert the temperatures to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 21 °C + 273.15 = 294 K

K = -48 °C + 273.15 = 225 K

Step 3: Calculate the final volume of the balloon

We will use the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂/ T₁ × P₂

V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr

V₂ = 1.27 × 10⁵ L

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

Explanation:

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

Answer:

Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).

Answer:

irreversible.

I hope this will help you

Answer:

The pH is greater than 7 at the equivalence point.

Explanation:

Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.

When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.

Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.

A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.

Weak acids are acids which only ionize partially in aqueous solutions.

When weak acids are dissolved in water, they produce only few hydrogen ions.

A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.

The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.

Learn more about equivalence point at: https://brainly.com/question/18933025

Answer:

Cu+(aq)--->Cu2+(aq) + e- : oxidation

reason: there is loss of electrons.

I2(s) + 2e--->2I-(aq) : reduction

reason: There is reduction of electrons.

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

Answer:

I do not speak Spanish.

Explanation:

Answer:

Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.

Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.  

The nucleophile in these reactions are new and called enols and enolates.

Explanation:

The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.  

Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.  

Aldehyde hydrogens not given Greek leters.  

α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.  

Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.  

The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.  

Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.  

The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.  

The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.

The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.

Aromaticity can also stabilize the enol tautomer over the keto tautomer.

Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.  

Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

The structures of both acetone and propanal are shown below:

In the formula of propanal there is -CHO functional group at the end.

In acetone -CO- group is present in the middle that is on the second carbon.

The molecular formula is C3H6O.

Both have same molecular formula but different structural formulas.

Answer:

no

Explanation:

Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.

Answer:

Explanation:

CCl4 => C + 2Cl2

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders