Which of these statements about enzymes is NOT true? - If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor.
- When [S] << Km, the reaction is second order and V0 depends on [S] and [Et].
- Their kcat is a second order rate constant.
- The lower their Km, the better they recognize their substrate, but the lower their reaction rate.
- When [S] << Km, V0 depends on [S] and [Et].

Answers

Answer 1

Answer:

Answer:

1. True. 2. True. 3. Not true. 4. True. 5. True

Explanation:

1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).

2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.

3. No, because it is measured in sec-1 and that means it is 1 rate constant.

4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.

5. The same as question two.

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Which type of bond is most likely to occur in the compound mgcl2?

Answers

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A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is thelimiting reactant and how much excess reactant remains after the reaction
has stopped?

Answers

Answer:

Ammonia is limiting reactant

Amount of oxygen left = 0.035 mol

Explanation:

Given data:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Excess reactant's amount left = ?

Solution:

Balance chemical equation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

NH₃ : N₂

4 : 2

0.12 : 2/4×0.12 = 0.06

NH₃ : H₂O

4 : 6

0.12 : 6/4×0.12 = 0.18

O₂ : N₂

3 : 2

0.125 : 2/3×0.125 = 0.08

O₂ : H₂O

3 : 6

0.125 : 6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

NH₃ : O₂

4 : 3

0.12 : 3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left = 0.125 - 0.09 = 0.035 mol

Final answer:

The limiting reactant in the reaction between Ammonia and Oxygen is Ammonia (NH3). All of the Oxygen is used up in the reaction, so no excess reactant remains.

Explanation:

This question involves a concept in chemistry known as limiting reactants and stoichiometry. The balanced chemical reaction between Ammonia (NH3) and Oxygen (O2) is: 4NH3 + 5O2 -> 4NO + 6H2O. This indicates that 4 moles of NH3 react with 5 moles of O2.

To find the limiting reactant, you first need to convert the grams of your reactants to moles. The molar mass of NH3 is approximately 17.0g/mol, and the molar mass of O2 is 32.0g/mol. Therefore, you have 2.00g/17.0g/mol = 0.118 moles of NH3 and 4.00g/32.0g/mol = 0.125 moles of O2.

Since 5 moles of O2 are needed for every 4 moles of NH3, and we have slightly more O2 than NH3, the limiting reactant is NH3. To find the amount of excess reactant, we determine how much O2 actually reacted by multiplying (0.118 moles NH3)*(5 moles O2/4 moles NH3) = 0.1475 moles O2. The original amount of O2 was 0.125 moles, so the amount left over is 0.125 - 0.1475, which is a negative number and thus not possible. This confirms that O2 is the excess reactant, although it entirely reacted. Hence, no excess reactant remains.

Learn more about Limiting Reactant here:

brainly.com/question/33417913

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When a substance changes its state, all BUT ONE of these statements istA)
Mass remains constant.
B)
Only the form changes.
Kinetic energy changes.
D)
Something chemically new is formed.
Science

Answers

Answer:

"Something chemically new forms" is the incorrrect statement

Explanation:

When a substance changes its state, it retains its mass even though its form and kinetic energy changes. It is simply moving between states, that is: loosing energy or gaining it. its chemical composition does not change.

Have a good day dude

Answer:

Science...Only the form changes.Kinetic energy changes.

Explanation:

SCIENCE

In polar coordinates, the origin is called the _____.

Answers

In polar coordinates, the origin is called the pole. Polar coordinate system is actually a two dimensional coordinate system. In this type of system the reference point is known by the name of pole. I hope that this is the answer that you were looking for and the answer has come to your desired help.

Answer:

In polar coordinates, the origin is called the pole.

Explanation:

The position of a point in a polar coordinate system is defined by using the distance from a fixed point and an angle made with a fixed ray from that point. It is a two-dimensional coordinate system.

The polar coordinates is defined as

$P(r,\theta)$

Where, r is radius and θ is the angle with the polar axis, i.e. positive axis.

In a coordinate system (0,0) is called origin.

In polar coordinate system (0,0) is called pole.

Therefore, in polar coordinates, the origin is called the pole.

What is the shortest day of the year known as? A.vernal B.equinox C.winter D.solstice summer solstice E.autumnal equinox

Answers

Answer:

summer solstice

Explanation:

on june solstice the northen hemisphere leans more toward the sun giving us longer days and more strong sunlight and It's the opposite in the Southern Hemisphere where june marks the start of winter thus its the shortest day of the year

200 mL of nitrogen gas at 24K is heated to 47K.

Answers

Answer:

See explanation

Explanation:

The question is incomplete. However, base on the given information, one likely question could be that, we calculate the the volume of the nitrogen gas after heating it.

So, we have that:

Initial Volume, V1 = 200mL

Initial Temperature, T1 = 24k

Final Temperature, T2 = 47k

To solve the final volume of the gas, we apply the following ideal gas law equation

V1 * T1 = V2 * T2

Substitute values for V1, T1 and T2

200mL * 24K = V2 * 47K

200mL * 24 = V2 * 47

4800mL = 47V2

Solve for V2

V2 = 4800/47