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Amiraneli [1.4K]
3 years ago
8

Alpha-mannosidase follows Michaelis-Menten kinetics. A 3 micromolar solution of alpha-mannosidase catalyzes the formation of .66

moles of p nitrophenol is-1 when incubated with the substrate 4-nitrophenyl alpha D-mannoside at a concentration 130 times greater than the Km for that substrate. What is alpha mannosidase Km
Chemistry
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

220

Explanation:

V₀= Vmax(D)/ (D) + km

0.66/100=vmax(130km)/(130km)+(km)

Vmax= 0.66×131/130×1000 =0.6650mm/d

kwat= vmax/ET

=0.665×10⁻³/ 3×10⁻⁶

= 0.22×10³

= 220

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Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE
allochka39001 [22]

Answer:

true

Explanation:

5 0
3 years ago
c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a
egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0
3 years ago
Hen a wire carrying a current is wound into a series of loops called coils, each turn of the wire produces a magnetic field. The
Lady_Fox [76]

Answer:

<u><em>Hope it helped </em></u>↓

Explanation:

Electromagnets are basically coils of wire which behave like bar magnets with a distinct north and south pole when an electrical current passes through the coil. The static magnetic field produced by each individual coil loop is summed with its neighbor with the combined magnetic field concentrated like the single wire loop we looked at in the last tutorial in the center of the coil. The resultant static magnetic field with a north pole at one end and a south pole at the other is uniform and a lot more stronger in the center of the coil than around the exterior. So a good question could be "Are electromagnets a magnet that respells, or attracts?" Then find away to explain that question/

(you don't have to do this question it is just a example.)

4 0
3 years ago
Read 2 more answers
At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the ΔH o f values of the reactant and pr
Xelga [282]

Answer:

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

Explanation:

<u>Step 1:</u> Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = -635.6 kJ/mol

ΔH°f (CO2) = -693.5 kJ/mol

ΔH°f (CaCO3) =-1206.9 kJ/mol

<u>Step 2: </u>The balanced equation

CaCO3(s) → CaO(s) + CO2(g)

<u>Step 3:  </u>Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) -  ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = -122.2 kJ /mol

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

<u>Step 5:</u> Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = -189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

7 0
3 years ago
If 12.8 g of CaCO3 decomposes at 38 degrees C and 0.96 atm, how many dm3 of CO2 are formed in addition to CaO? CaCO3 → CaO + CO2
jasenka [17]

Answer:

3.4dm3

Explanation:

We'll begin by calculating the number of mole of CaCO3 present in 12.8g of CaCO3.

This can be achieved as shown below:

Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

Mass of CaCO3 obtained from the question = 12.8g

Number of mole of CaCO3 =?

Number of mole = Mass /Molar Mass

Number of mole of CaCO3 = 12.8/100

Number of mole of CaCO3 = 0.128 mole

The equation for the reaction is given below:

CaCO3 → CaO + CO2

From the equation above,

1 mole of CaCO3 produced 1 mole of CO2.

Therefore, 0.128 mole of CaCO3 will also produce 0.128 mole of CO2.

Now, we can obtain the volume of CO2 produced as follow:

Data obtained from the question include:

T (temperature) = 38°C = 38 + 273 = 311K

P (pressure) = 0.96 atm

n (number of mole of CO2) = 0.128 mole

R (gas constant) = 0.082atm.dm3/Kmol

V (volume of CO2) =?

Using the ideal gas equation PV = nRT, the volume of CO2 produced can be obtained as shown:

PV = nRT

0.96 x V = 0.128 x 0.082 x 311

Divide both side by 0.96

V = (0.128 x 0.082 x 311) /0.96

V = 3.4dm3

Therefore, 3.4dm3 of CO2 are produced.

7 0
3 years ago
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