N=5.0 mol
m=3.2 kg
C=n/m
C=5.0/3.2≈1.6 mol/kg
Plant cells have chloroplasts and a cell wall..... chloroplasts are the organelle that conduct photosynthesis..... and the cell wall is like an extra barrier for the cell membrane.... I hope this helps, have an amazing day/or night.... and god bless uuuu!!!
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Answer:
1.86% NH₃
Explanation:
The reaction that takes place is:
- HCl(aq) + NH₃(aq) → NH₄Cl(aq)
We <u>calculate the moles of HCl that reacted</u>, using the volume used and the concentration:
- 32.27 mL ⇒ 32.27/1000 = 0.03227 L
- 0.1080 M * 0.03227 L = 3.4852x10⁻³ mol HCl
The moles of HCl are equal to the moles of NH₃, so now we <u>calculate the mass of NH₃ that was titrated</u>, using its molecular weight:
- 3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃
The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:
- 0.0592 / 12.949 * 100% = 0.4575%
Now we <u>calculate the total mass of NH₃ in the diluted sample</u>:
Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g
- 0.4575% * 95.806 g = 0.4383 g NH₃
Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:
- 0.4383 g NH₃ / 23.495 g * 100% = 1.86% NH₃
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change 
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
Thus, we obtain a quadratic equation whose solution is:
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
