P1V1=P2V2
(1.00atm)(3.60L)=(2.50atm)V
3.60=(2.50atm)V
3.60/2.50=V
1.44L=V
Answer:
If you are given a chemical equation and specific amounts for each reactant in grams, you have to follow these steps, in order, to determine how much product can possilby be made:
1. Convert each reactant into moles of the product.
2. Determine which reactant is the limiting reactant.
3. Convert the moles of product, from the limiting reactant, to grams.
Explanation:
Answer:
Explanation:
1)
HH H H H
| | | | |
H- C-C-C-C-C-H
| | | | |
H H H H H
This one can be written as
CH3 - CH2 - CH2 - CH2 - CH3
2)
H H H H
| | | |
H -C- C-C- C-H
| | | |
H H | H
H -C-H
|
H
This one can be written as
CH3 - CH2 -CH -CH3
|
CH3
The concentration of the solution is 8 M
<u><em>calculation</em></u>
<em> </em>Concentrati<em>on = moles/volume in liters</em>
<em> </em><em>step 1: find moles of HF</em>
<em>moles of HF =mass/molar mass</em>
<em>molar mass of HF = 1+ 19 )= 20 g/mol</em>
<em> moles is therefore = 32.0 g/ 20 g/mol= 1.6 moles</em>
<em>Step 2: convert ml to L</em>
<em>volume in liters = 2.0 x 10^2 / 1000 =0.2 l</em>
<em>step 3: find the concentration</em>
<em>concentration = 1.6 mol / 0.2 l = </em><em>8 M</em>
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + <u>2H⁺ + 2e⁻</u>; E°´ = -0.031 V
<u>NAD⁺ + </u><u>2H⁺ + 2e⁻</u><u> ⇌ NADH + H⁺; </u> E°´ = <u> -0.320 V</u>
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
<u>FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; </u> E°´ = <u>-0.219 V
</u>
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
