Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
Answer:
- Alanine = 5.61 mmoles
- Leucine = 3.81 mmoles
- Tryptophan = 2.45 mmoles
- Cysteine = 4.13 mmoles
- Glutamic acid = 3.40 mmoles
Explanation:
Mass / Molar mass = Moles
Milimoles = Mol . 1000
500 mg / 1000 = 0.5 g
- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles . 1000 = 5.61mmoles
- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles . 1000 = 3.81 mmoles
- Tryptophan = 0.5 g / 204 g/m → 2.45x10⁻³ moles . 1000 = 2.45 mmoles
- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles
- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 = 3.4 mmoles
9.184 liters CH2O at STP
I think this is correct. Good luck
Answer:
The minimum concentration of acetaminophen that can be detected by new= 10μg/mL
Volume of blood sample=2ml
Minimum mass of acetaminophen that can be detected by automated system= 10×2= 20μg= 0.020mg