Answer:
A 12 oz Coca Cola contains 39g of sugar or C6H12O6.
To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:
39g/ 180.16 g/mol = 0.216 mol sugar
then, convert 12 oz to L:
12oz / (1oz/0.02957L) = 0.35484 L
therefore the concentration of sugar in the soda is:
M = mol sugar / L sol'n
= 0.216 mol sugar / 0.35484 L
= 0.609 M
Explanation:
To name this Alkyne, simply count from the direction that will give the lowest starting number to appear at the beginning of the carbon triple bond.
If you were to count from the top of the chain, the position of the carbon next to the triple bond would be 4. Yet if you count from the bottom chain going left to right and above the chain, the position of the carbon next to the triple bond would be 3.
Then identify the groups that are connected off the parent chain, here we have a methyl group on carbon 2.
Thus the name would be 2 - methyl - 3 - heptyne. I believe.
Answer:
The answer to the question is
<h2>
A. Air, Soil, Sand, Water</h2>
Explanation:
Explanation:
where,
R = Gas constant = 
T = temperature = ![600^oC=[273.15+600]K=873.15 K](https://tex.z-dn.net/?f=600%5EoC%3D%5B273.15%2B600%5DK%3D873.15%20K)
= equilibrium constant at 600°C = 0.900
Putting values in above equation, we get:
The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.
Equilibrium constant at 600°C = 
Equilibrium constant at 1000°C = 
![T_1=[273.15+600]K=873.15 K](https://tex.z-dn.net/?f=T_1%3D%5B273.15%2B600%5DK%3D873.15%20K)
![T_2=[273.15+1000]K=1273.15 K](https://tex.z-dn.net/?f=T_2%3D%5B273.15%2B1000%5DK%3D1273.15%20K)
![\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7B%5CDelta%20H%5Eo%7D%7BR%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
![\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7B0.396%7D%7B0.900%7D%3D%5Cfrac%7B%5CDelta%20H%5Eo%7D%7B8.314%20J%2Fmol%20K%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B873.15%20K%7D-%5Cfrac%7B1%7D%7B1273.15%20K%7D%5D)
The ΔH° of the reaction at 600 C is -18,969.30 J/mol.
ΔG° = ΔH° - TΔS°
764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°
ΔS° = -22.60 J/K mol
The ΔS° of the reaction at 600 C is -22.60 J/K mol.
Partial pressure of carbon dioxide = 
Partial pressure of carbon monoxide = 
Where 
mole fraction of carbon dioxide and carbon monoxide gas.
The expression of is given by:
Mole fraction of carbon dioxide at 600°C is 0.474.
The compound is CH₃CH₂CH₂NH₂
when naming the compound the C chain has 3 C atoms and they all are in a single chain therefore alkane chain is propane.
It has a -NH₂ group attached to the C therefore functional group is 'amine'
-NH₂ is attached to a terminal C atom therefore when numbering the C atoms the C attached to the substituent should have the lowest number therefore C atom with NH₂ is carbon atom 1
therefore name of the compound is 1-propanamine
