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
That would be telophase 2 of Meiosis
Answer:
Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.
Answer:
2.8M
Explanation:
The following data were obtained from the question:
Volume of stock solution (V1) = 40mL
Molarity of the stock solution (M1) = 7M
Volume of diluted solution (V2) = 100mL
Molarity of diluted solution (M2) =?
Using the dilution formula, we can easily find the molarity of the diluted solution as follow:
M1V1 = M2V2
7 x 40 = M2 x 100
Divide both side by 100
M2 = (7 x 40)/100
M2 = 2.8M
Therefore, the molarity of the diluted solution is 2.8M
The answer for this question is:
B. Scientific notation is used to keep track of very small and very large numbers during mathematical calculations.