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Masteriza [31]
3 years ago
5

Enolase, an enzyme that catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate: Group of answer choices demonstra

tes a base-catalysis step, where a proton is removed from the substrate. demonstrates an acid-catalysis step, where a proton is added to the substrate. requires two Mg2 cofactors and as such demonstrates metal-ion catalysis. stabilizes the intermediate and transition
Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

Requires two Mg²⁺ cofactors and as such demonstrates metal-ion catalysis

Explanation:

Electrostatic catalysis or metal ion catalysis is a catalytic mechanism that makes use of metalloenzymes, such as enolase along with a metal ion which is bound tightly, including, Mo⁶⁺, Ni³⁺, Co³⁺, Mn²⁺, Zn²⁺, Cu²⁺, and Fe²⁺, to undertake a catalysis

For maximal activity, enolase requires the presence of 2 equivalent metal ions in each site which is active

Therefore, the correct option is; requires two Mg²⁺ cofactors and as such demonstrates metal-ion catalysis

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How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0℃.
Ivanshal [37]

Answer:

There will react 30.9 milligrams of magnesium

Explanation:

Step 1: Data given

Volume of hydrogen = 31.2 mL

Pressure = 754 torr = 754/760 = 0.992 atm

Temperature = 25.0 °C = 298 Kelvin

Step 2: The balanced equation

Mg + 2HCl → MgCl2 + H2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 0.992 atm

⇒with V = the volume of H2 = 31.2 mL = 0.0312 L

⇒ with n = the moles of H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 25.0 °C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.992*0.0312)/(0.08206*298)

n = 0.00127 moles

Step 4: Calculate moles of Mg

For 1 mol of Mg we need 2 moles of HCl to produce 1 mol of MgCl2 and 1 mol of H2

For 0.00127 moles of H2 we need 0.00127 moles of Mg

Step 5: Calculate mass of Mg

Mass of Mg = moles of Mg * molar mass of Mg

Mass of Mg = 0.00127 moles * 24.3 g/mol

Mass of Mg = 0.0309 grams = 30.9 mg of Mg

There will react 30.9 milligrams of magnesium

3 0
3 years ago
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.
kherson [118]

Answer:

The partial pressure of the other gases is 0.009 atm

Explanation:

Step 1: Data given

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.

The atmospheric pressure = 0.90 atm

Step 2: Calculate mol fraction

If wehave  100  moles of air, 78 moles will be nitrogen,

21  moles will be  oxygen, and  1  mol will be other gases.

Mol fraction = 1/100 = 0.01

Step 3: Calculate the partial pressure of the other gases

Pgas = Xgas * Ptotal

⇒ Pgas = the partial pressure = ?

⇒ Xgas = the mol fraction of the gas = 0.01

⇒Ptotal = the total pressure of the pressure = 0.90 atm

Pgas = 0.01 * 0.90 atm

Pgas = 0.009 atm

The partial pressure of the other gases is 0.009 atm

4 0
3 years ago
An element has the mass number 12 and atomic number 6. The number of neutron in it is?
slamgirl [31]

Answer:

12-6=6

the answer is A)6

8 0
3 years ago
Chemical reactions that ____ energy will not occur without a source of energy
andrezito [222]
Chemical reactions that release energy will not occur without a source of energy.  So the answer is release.
4 0
3 years ago
A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate
Tomtit [17]

A.       The radioactive decay equation is N = N0e^{-ln(2)*t/T }

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = e^{-ln(2)*t/T }

Since N = (1/8) N0 and substituting known values:

1/8 = e^{-ln(2)*t/5730}

Taking ln of both sides:

ln(1/8)= -ln(2)*t/5730

t = - 5730 * ln(1/8) / ln (2)

t = 17,190 years

The tree was cut down 17,190 years ago.

B.   N0 = 1,500,000 carbon-14 atoms

Since N = (1/8) N0

N = 187,500 carbon atoms left

3 0
3 years ago
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