An ionic compound is composed of ionic bonds that are formed by transfer of electrons from one atom to the other. The atom that loses electrons acquires a positive charge (cation) while that which gains electrons acquires a negative charge.
In the case of sodium chloride; Sodium Na has 1 electron in its outer orbital while Chlorine Cl has 7 electrons. Thus, Cl requires 1 electron to complete its octet. This electron is donated by Na.
Thus, NaCl is essentially, Na⁺Cl⁻
Ans D) Chlorine becomes an anion by gaining an electron from sodium
The answer is 2. Temperate.
For the pH of a solution prepared by dissolving 0. 350 mol of solid methylamine hydrochloride, the pH is mathematically given as
pH = 11.14
<h3>What is the pH value of a solution prepared by dissolving 0. 350 mol ?</h3>
Generally, the equation for the pKb is mathematically given as
pKb = -logKb
Therefore
pKb of CH3NH2 = -log(4.40×10-4)
pKb of CH3NH2= 3.36
Hence
pKa= 14 - pKb
pKa= 14 - 3.36
pKa= 10.64
In conclusion, using
pH = pKa + log([A-]/[HA])
pH = 10.64 + log(1.10M/0.350M)
pH = 11.14
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brainly.com/question/1584537
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
