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exis [7]
2 years ago
6

What is the theoretical yield of moles of hydrogen that can be produced from 0.032g of MG?

Chemistry
1 answer:
Ugo [173]2 years ago
5 0

Answer:

1.31x10⁻³ moles of H₂

Explanation:

This is the equation:

Mg(s)  +   2H₂O (g)   →   Mg(OH)₂ (aq)   +    H₂(g)

Ratio is 1:1, so 1 mol of Mg is needed to produce 1 mol of H₂

Mass / Molar mass = Mol

0.032 g / 24.3 g/m = 1.31x10⁻³ moles

1.31x10⁻³ moles of H₂(g)

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3 years ago
The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?
Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃          101.96 g
------------------------  x  -------------------  = 275 g Al₂O₃  = 280 g Al₂O₃
                                     1 mole

5 0
2 years ago
An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
fomenos

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

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2 years ago
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