The mass of oxygen is 236.4 g.
Explanation:
To find the mass of oxygen in 325 g of carbon dioxide (CO₂) we follow the next algorithm.
First we calculate the molar weight of carbon dioxide:
molar weight of CO₂ = molar weight of carbon × 1 + molar weight of oxygen × 2
molar weight of CO₂ = 12 × 1 + 16 × 2 = 44 g/mole
Now we devise the following reasoning:
if in 44 g of CO₂ there are 32 g of oxygen
then in 325 g of CO₂ there are X g of oxygen
X = (32 × 325) / 44 = 236.4 g oxygen
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Answer: There are 6 different outcomes.
Explanation:
There are two options for procedure one (one or two steps) and three options for procedure two (one, two or three steps). If we multiply those options , the result is the total number of outcomes for the analysis.
2 x 3 = 6
These being
- One step on procedure one and one step on procedure two
- One step on procedure one and two steps on procedure two
- One step on procedure one and three steps on procedure two
- Two steps on procedure one and one step on procedure two
- Two steps on procedure one and two steps on procedure two
- Two steps on procedure one and two steps on procedure two
Explanation:
1)  + 7 H_2(g)](https://tex.z-dn.net/?f=%202%20Al%28s%29%20%2B%202%20NaOH%28aq%29%20%2B%206%20H_2O%28l%29%20%5Clongleftrightarrow%202%20Na%5BAl%28OH%29_4%5D%28aq%29%20%2B%207%20H_2%28g%29)
![Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNa%5BAl%28OH%29_4%5D%5D%5E2%2A%5BH_2%5D%5E7%7D%7B%5BNaOH%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

2) 
![Kc=\frac{[H_2SO_4]}{[SO_3]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2SO_4%5D%7D%7B%5BSO_3%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

3)
![Kc=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

That you have to do better
Answer:
Explanation:
Percent composition is percentage by the mass of element present in the compound.
Given , Mass of sulfur= 32.1 amu
Mass of oxygen = 16.0 amu
Mass of sulfur trioxide
= 32.1 amu + 3*16.0 amu = 80.1 amu