Answer:
191.11 grams of oxygen gas should be produced.
Explanation:
The balanced reaction is:
2 Al₂O₃ → 4 Al + 3 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- Al: 4 moles
- O₂: 3 moles
Being the molar mass of each compound:
- Al₂O₃: 102 g/mole
- Al: 27 g/mole
- O₂: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles* 102 g/mole= 204 grams
- Al: 4 moles* 27 g/mole= 108 grams
- O₂: 3 moles* 32 g/mole= 96 grams
Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

mass of oxygen= 191.11 grams
<u><em>191.11 grams of oxygen gas should be produced.</em></u>
You can cross multiply and then get k alone so it would look like this:
76 81
----- = ------ = 76k= 24624
304 k ----- ----------- = k = 324
76 76
so k is equal to 324
hope this helps have a nice day
Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!
Answer:
2H2S + 3O2 → 2SO2 + 2H2O
Explanation:
Step 1: Data given
Hydrogen sulfide = H2S
Oxygen = O2
sulfur dioxide = SO2
water = H2O
Step 2: The unbalanced equation
H2S + O2 → SO2 + H2O
Step 3: Balancing the equation
H2S + O2 → SO2 + H2O
On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.
H2S + 3O2 → 2SO2 + 2H2O
On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.
Now the equation is balanced.
2H2S + 3O2 → 2SO2 + 2H2O
Because the 4s orbital is at a lower energy level.
The 4s orbital has a lower energy because it is more <em>penetrating</em> than a 3d orbital. A 4s electron can get closer the nucleus than a 3d electron, so it has a lower potential energy.
The other options are incorrect because they contradict the correct one.