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k0ka [10]
3 years ago
10

An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for

S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Tube number: 1
Rate of formation of product:0.5
Tube number: 2
Rate of formation of product: 4.8
Chemistry
1 answer:
Lera25 [3.4K]3 years ago
3 0

Answer:

100.5 ≈ 101

Explanation:

Km for S1 = 2.0 mM

Km for S2 = 20 mM

Given that : S1 = S2 =  hence Vmax for either S1 or S2 can represent

The Vmax can be calculated using the data Given and equation below

Vo = Vmax [s] / ( Km + [s] )  ------ (1)

Vo = 0.5

[s] = 0.1 mM

km = 20 mM

making Vmax subject of  equation 1

Vmax = 0.5 ( 20.1 mM) / (0.1 mM )

         = 100.5 ≈ 101

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guapka [62]

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

<u>Explanation:</u>

Oxidizing agent is defined as the chemical reagent which helps the other chemical compound to get oxidized and itself gets reduced. The oxidation state for these species gets reduced because they are undergoing reduction reaction.

For the given chemical equation:

K_2Cr_2O_7+14HCl\rightarrow 2CrCl_3+2KCl+3Cl_2+7H_2O

Oxidation state of Chromium is getting reduced from +6 to +3 and oxidation state of chlorine getting increased from -1 to 0.

Hence, K_2Cr_2O_7 acts like and oxidizing agent because it is itself getting reduced to CrCl_3

Therefore, the correct answer is Option b.

4 0
3 years ago
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
Which are characteristics of a prokaryotic cell? Select three options. contains DNA lacks DNA contains ribosomes lacks ribosomes
lara [203]
Contains DNA, contains ribosomes, lacks a nucleus
7 0
3 years ago
Read 2 more answers
Which atom, selenium (Se) or Arsenic (As), has the higher electron affinity and why?
IrinaVladis [17]

Answer:

5. Selenium, because it does not have a stable, half-filled p subshell and adding an electron does not decrease its stability.

Explanation:

Electron affinity is the amount of energy released when an isolated gaseous atom accepts electron to form the corresponding anion.

Selenium:-

The electronic configuration of the element is:-

[Ar]3d^{10}4s^24p^4

Arsenic:-

The electronic configuration of the element is:-

[Ar]3d^{10}4s^24p^3

The 4p orbital in case of arsenic is half filled which makes the element having more stability as compared to selenium.

Thus, selenium has higher electron affinity because adding electron does not decrease the stability as in case of arsenic.

8 0
3 years ago
Be sure to answer all parts. Write a balanced equation (including physical states) for the following reaction: Sodium carbonate,
aalyn [17]

Answer:

I think it would be:

NaCO3 (s)-->Na2O (s) + CO2 (g)

3 0
3 years ago
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