Answer:
An input of heat energy Yr don't you know stupied
The answer to this question would be: <span>NaOH solution
Limiting reagent is the reagent that totally consumed in the reaction. To answer this question you need to know the equation for the chemical reaction. The equation would be:
</span><span>HCl + NAOH = H2O + NACl
</span>
All of the coefficient in the reaction is 1, which mean 1 HCl will react to 1 NaOH. In this question, the concentration of HCl and NaOH is same so the amount of the molecule can be reflected by the volume. NaOH has lower volume compared to HCl so it is clear that NaOH will be totally consumed in this reaction.
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
Answer
Manganese II oxide
Thank you
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:
Making Ki subject of the formula:
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
