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!
Answer: Fe<em>(aq)</em>+S<em>(aq)</em>=FeS<em>(s)</em>
Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:
1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)
Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:
Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr
2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.
3.) Ionic formula: Fe(aq)+ <u>2Br(aq)</u>+<u>2 Na(aq)</u>+S (aq)=FeS(s)+<u>2 Na(aq)+2Br(aq)</u>
4.)When you've derived a total ionic equation (above), you'll find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)
Hope this helps :)
The theory is most likely accurate and the main reason behind this process of thinking is because a majority of the scientists around the world agree with the theory. It can never be biased or incomplete. The correct option among all the options that are given in the question is the third option or option "C".
Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
=
u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
Explanation:
The dimensions of a standard backpack is 30cm x 30cm x 40cm
The mass of an average student is 70 kg
We know that, the density of gold is 19.3 g/cm³.
Let m be the mass of the backpack. So,

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.