Https://wellnessmama.com/26519/risks-essential-oils/ is what u can learn maybe u can find ue answer Hope i helped best of luck :)
Answer:
A/1. 10.9 mol O2
Explanation:
583 g x 1 mol SO3 x 3 mol O2 /
80.057 g mol SO3 x 2 mol SO3
- You just need to find molar mass of SO3, which is 80.057 g.
- Everything else came from formula. Further explanation...
- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.
- If you need a further explanation or more help on any problems I would be happy to help, just let me know.
Explanation:
Given parameters:
Number of molecules = 4.21 x 10²³ molecules
Unknown parameters:
Number of moles
Solution:
A mole can be defined as the amount of a substance that contains the avogadro's number of particles i.e 6.02 x 10²³
To find the number of moles:
Number of moles = ![\frac{number of molecules }{[tex]6.02 x 10^{23}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20of%20molecules%20%7D%7B%5Btex%5D6.02%20x%2010%5E%7B23%7D)
}[/tex]
Number of moles = ![\frac{4.21 x 10[tex]^{23}](https://tex.z-dn.net/?f=%5Cfrac%7B4.21%20x%2010%5Btex%5D%5E%7B23%7D)
}{
}[/tex]
Number of moles of CaCl₂ = 0.699moles
Learn more;
mole calculation brainly.com/question/13064292
#learnwithBrainly
Direct Proportion and The Straight Line Graph.Straight<span> line graphs that go through the origin, like the one immediately below, show that the quantities on the graph are in direct proportion.
thanks
cbuck763
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Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
