Ethanol (CH3CH2OH) is the alcohol found in beverages. It is oxidized in the body to

Chemistry

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

B) Ethanol must act as a competitive inhibitor for the alcohol dehydrogenase and therefore slows the formation of formaldehyde.

Explanation:

An<em> inhibitor</em> is a substance that binds to an enzime and prevents it to react with a substrate. In this case, the inhibitor is ethanol, binding to the enzime (alcohol dehydrogenase) <u>and preventing it to react with methanol and thus from it form the toxic formaldehyde</u>.

A competitive inhibitor has a similar structure to the substrate, that is why it binds to the same active site. In this case, ethanol has a similar structure to methanol, which is why the answer is B and not A.

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Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m

faust18 [17]

Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

The interplanar spacing can be calculated from:

$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

$\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476$

Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

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$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$

4 0

3 years ago

72.0 grams of water how many miles of sodium with react with it?

Flura [38]

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$