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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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When you multiply measurements, the answer should only have the same number of significant figures as the measurement with the _
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Answer: fewest

Explanation:

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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

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We cannot see it its so small

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Calculate average kinetic energy of one mole of gas at 517K
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