The amount of sugar is 2621 mg
Why?
The complete question is:
A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.
Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?
To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:

Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:

Have a nice day!
Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
![pH = -log([H_{3}O^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20)
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
<u>Trimethyl ammonium</u>:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
![Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5B%28CH_%7B3%7D%29_%7B3%7DN%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5B%28CH_%7B3%7D%29_%7B3%7DNH%5E%7B%2B%7D%5D%7D)

By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
<u>Phenol</u>:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
![Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DOH%5D%7D)


Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%289.99%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.00%20)
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
Therefore, the difference in pH values is 4.95.
I hope it helps you!
Well, we need to find the ratio of Al to the other reactant.
Al:HCl = 1:3
--> this means that for every 1 Al used, you have to use 3 HCl.
6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.
The ratio of HCl:AlCl = 3:1
13/3 = 4.3333...
The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.
Hope this helps!!! :)
To balance the following chemical equation, make a tally or a count of each of the atoms on both sides of the reaction, and make sure that those atoms are equal on both the reactant and product side.
AL2O3 + HCl => ALCl3 + H2O
Left side. Right side
AL = 2. AL = 1
O = 3 Cl = 3
H = 1. H = 2
Cl = 1. O = 1
First balance the metal atoms, aluminum, then hydrogen and then oxygen.
Balanced equation :
AL2O3 + 6HCl => 2ALCl3 + 3H2O.
Left side. Right side
AL = 2 AL = 2
O = 3 Cl = 6
H = 6 H = 6
Cl = 6 O = 3.