It should be 8 O atoms. 3O atoms in Na2S2O3 and 5O atom in 5H2O. The reason there are 5 O atoms are because the 5 in front of H2O means you multiply each atom in the compound by that number (like the distributive property). The H2 molecule becomes 10 Hydrogen atoms (5*2) and the Oxygen becomes 5 Oxygen atoms (5*1). Then you add the 5O atoms to the 3O atoms which equals 8
Helium is the lowest boiling point.
The molarity of the solutions are as follows:
- solution B has the highest molarity
- solutions A, D and F have the same molarity
- solutions A and C are mixed together have a lower molarity than B
- solution F and D will have the same molarity
- Volume of water required to be evaporated is 8.3 mL
<h3>What is molarity of a solution?</h3>
The molarity of a solution is the amount in moles of a substance present in a given volume of solution.
From the image of the solution given:
- solution B has the highest molarity
- solutions A, D and F have the same molarity
- when solutions A and C are mixed, the resulting solution have a lower molarity than B
- solution F and D will have the same molarity after 75 mL and 50 mL of water are added to each respectively
- the molarity of B is 12/50 = 4/16.7. Volume of water required to be evaporated = 25 - 16.7 = 8.3 mL
Therefore, the molarity of the solutions depends on the moles of substance present per given volume of solution.
Learn more about molarity at: brainly.com/question/24305514
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Answer:
2,7 m
Explanation:
You can solve this doing an energy balance:
![m*g*h-\frac{1}{2} *m*v^{2} =41,7[J]](https://tex.z-dn.net/?f=m%2Ag%2Ah-%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%3D41%2C7%5BJ%5D)
Solving this equation to get h:
![\frac{41,7- \frac{1}{2} *m*v^{2} }{m*g}=h](https://tex.z-dn.net/?f=%5Cfrac%7B41%2C7-%20%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%7D%7Bm%2Ag%7D%3Dh)
Replacing the values and solving brings to 2,7 m
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>