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Troyanec [42]
2 years ago
6

Which of the following describes the effect of boiling and freezing? which of the following describes the effect of boiling and

freezing? boiling and freezing both denatured amylase. boiling denatured the enzyme, but freezing had no effect. freezing denatured the enzyme, but boiling had no effect. boiling and freezing had no effect on amylase?
Chemistry
1 answer:
Natali5045456 [20]2 years ago
7 0

Boiling denatures the enzyme, but freezing has no effect.

At high temperatures (even if it’s not boiling, about 60°C is enough) hydrogen bonds and other interactions between different amino acids of an enzyme are broken. This is not degradation, no bonds are broken – only interactions (London, Van der Waals, etc). This makes the protein to denature, which means it will lose its 3D shape and become linear – no longer globular: it only preserves its primary structure. This makes the enzyme to become useless: it can’t make the substrate become product anymore. So the activity becomes practically zero.

Freezing DOES make a change in enzyme activity, but for other reasons which are not denaturing (the enzyme has the same structure as before):

-Concentration: Now some of the water has become ice and you have 2 phases, which are the solution and the ice. So the solution has less water, and so a higher concentration. Higher concentration means… higher activity! So it has the same structure, but it works more.

-Cell disruption: Membranes are broken and this will release molecules that used to be stored in an organelle. Sometimes this makes the enzyme find the substrate – so it actually increases activity! As I said for concentration, this means the enzyme has the same structure but it works more.


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A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
lilavasa [31]

Answer:

m_{Fe}=12.6gFe

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

Best regards.

3 0
2 years ago
Read 2 more answers
choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---> Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-
kompoz [17]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Standard enthalpy of formation is the change in enthalpy of  one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula Ca(NO_3)_2

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of Ca(NO_3)_2 from the components in their standard states follows:

Ca+N_2+3O_2\rightarrow Ca(NO_3)_3

Hence, the correct answer is Option A.

3 0
2 years ago
In science class, Blaine’s teacher puts one glow stick in a cup of hot water and another glow stick in a cup of cold water. She
timofeeve [1]

Answer:

The glow stick in hot water will be brighter

Explanation:

The glow stick in hot water will be brighter than the glow stick in cold water because the heat from the hot water will cause the molecules in the glow stick to move faster. The faster the molecules move in the glow stick, the sooner and brighter the reaction will be. The cold water will cause molecules to move slowly and it will take longer for the reaction to occur, which will also make it less bright.

3 0
2 years ago
Read 2 more answers
A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2
Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

  • The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

  • Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
  • van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0
2 years ago
What is the conjugate acid of each of the following? What is the conjugate base of each?
Lilit [14]

Answer:

a. H₂O (conjugate acid) ; b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid) ; c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base) ; d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base) ; f. No conjugate acid either base;  g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a.  OH⁻  +  H⁺  ⇄ H₂O

The hydroxide acts like a Bronsted Lory base, so it can catch a proton. Water will be the conjugate acid.

b. H₂O, is an amphoterus compound. It can act as an acid or a base. If it is a base, the conjugate acid is the H₃O⁺. If it is an acid, the conjugate base is the OH⁻.

c. HCO₃⁻  +  H⁺  ⇄  H₂CO₃

HCO₃⁻  +  H₂O  ⇄ CO₃⁻²  +  H₃O⁺

The bicarbonate is also amphoteric. When it catches the proton, the carbonic acid is the conjugate acid, cause it works as a base.

When the HCO₃⁻ (acid) release the proton, the conjugate base is the carbonate.

d. Ammonia is a weak base, so the conjugate strong acid is the ammonium.

NH₃ + H₂O  ⇄  NH₄⁺  +  OH⁻

e. Another amphoteric compound. The acid sulfate acts an acid and a base.

(like bicarbonate). When it is a base, the conjugate acid is the sulfuric acid, when it is an acid, the conjugate base is the sulfate.

HSO₄⁻  +  H₂O  ⇄  SO₄⁻²  +  H₃O⁺

HSO₄⁻  +  H⁺  ⇄  H₂SO₄

f. H₂O₂ does not recieve H⁺ or OH⁻, and it does not release H⁺. It is a neutral compound and it doesn't act as a base or acid.

g. HS⁻ is amphoterous.

HS⁻  +  H⁺  ⇄  H₂S

HS⁻  +  H₂O  ⇄  S⁻²  +  H₃O⁺

Same case as bicarbonate or acid sulfate.

h. H₅N₂⁺  +  H₂O  ⇄  H₄N₂  + H₃O⁺

Hidrazinium acts an acid, so, the conjugate base will be the hidrazine.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

3 0
2 years ago
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