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Gnoma [55]
3 years ago
13

At 900∘c titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chlor

ide. what is being oxidized, and what is being reduced?
Chemistry
2 answers:
german3 years ago
7 0
TiCl4 is being reduced and Mg being oxidized
liberstina [14]3 years ago
5 0

Answer:

Magnesium is being oxidized.

Titanium is being reduced.

Explanation:

Hello,

The undergoing chemical reaction is:

TiCl_4+2Mg-->Ti+2MgCl_2

Next, oxidation states are assigned:

Ti^{+4}Cl^-_4+2Mg^0-->Ti^0+2Mg^{+2}Cl^{-}_2

Half reactions are as follows:

Ti^{+4}+4e^--->Ti^0\\Mg^0-->Mg^{+2}+2e^-

Finally, as the magnesium increases its oxidation state, it is being oxidized and as the titanium decreases its oxidation state it is being reduced.

Best regards.

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3 years ago
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

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