Answer:
N2I6 = 789 g
N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
N2I6=0.136molesx789g/mole=107g=110g
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Hope this helps, have a BLESSED AND WONDERFUL DAY!
- Cutiepatutie ☺❀❤
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
First, multiply the mass by the molar mass of neon to find out how many moles of neon there are. Then, multiply by 22.4 to find out how many liters there are.
6.745g Ne x 1 mole Ne/20 g Ne x 22.4 L/1 mole Ne = 7.5544 L
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is 
Explanation:
From the question we are told that
the cell voltage for AD is 
From the data give we can see that

i.e 
In the same way we can say that

=> 

of octane had been converted to carbon dioxide CO₂.
<h3>Explanation</h3>
Octane has a molar mass of

1.000 gallon of this fuel would have a mass of 2.650 kilograms or
, which corresponds to
of octane.
Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

The mass of the product mixture is
heavier than that of the octane supplied. Thus
of oxygen were consumed in the combustion. There are
of oxygen molecules in
of oxygen.
Let the number of moles of octane that had undergone complete combustion as seen in the first equation be
(
). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal
.
25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.


Therefore
out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.
