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 is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
Molar mass Mg = 24.3 g/mol
1 mole mg ------------ 24.3 g
?? moles mg --------- 4.75 g
4.75 x 1 / 24.3 => 0.195 moles of Mg
hope this helps!
<span>Higher energy = shorter wavelength
Frequency is one cycle over an amount of time (seconds)
So higher frequency = higher energy = shorter wavelength</span>
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
