Search found 257 matches
- Sat Aug 20, 2022 7:49 am
- Forum: Chemistry forum
- Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
- Replies: 9
- Views: 3324
Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Author's given answer is wrong.The correct answer is + 1.21 volts using the formula for ∆G°= -nFE°
- Sat Aug 20, 2022 4:18 am
- Forum: Chemistry forum
- Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
- Replies: 9
- Views: 3324
Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Why did you choose ① as reduction half reaction and ② as oxidation half reaction?
Would you explain that?
Would you explain that?
- Sat Aug 20, 2022 4:02 am
- Forum: Chemistry forum
- Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
- Replies: 9
- Views: 3324
Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
I think your answer is correct. There is a typographical error in the given answer.
- Sat Aug 20, 2022 3:57 am
- Forum: Chemistry forum
- Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
- Replies: 9
- Views: 3324
Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Author has given the following answer:
is there typographical error in the above answer?- Sat Aug 20, 2022 2:17 am
- Forum: Chemistry forum
- Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
- Replies: 9
- Views: 3324
E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
It is sometimes necessary to compute E° for a given half reaction from other half reaction of known E°. For example, the standard electrode potential for the oxidation of Titanium metal to Ti³⁺ can be obtained from the following half reaction: Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts ① Ti³⁺ + e⁻ ⇌ Ti²⁺ E°...
- Thu Aug 18, 2022 10:58 pm
- Forum: Chemistry forum
- Topic: Solubility product constant
- Replies: 8
- Views: 8421
Re: Solubility product constant
I found on Internet, E°= 0.37 Volts for the reaction [Ag(NH₃)₂]⁺ + e⁻ ⇌ Ag + 2 NH₃.
So, K = Kf, Log Kf= \(\frac{0.80 Volts-0.37 Volts}{0.059} = 7.29 K_f= 1.9 \times 10^7 \)
There is a discrepancy between your answer of 2.1e6 and my answer 1.9e7. Is this discrepancy significant?
So, K = Kf, Log Kf= \(\frac{0.80 Volts-0.37 Volts}{0.059} = 7.29 K_f= 1.9 \times 10^7 \)
There is a discrepancy between your answer of 2.1e6 and my answer 1.9e7. Is this discrepancy significant?
- Thu Aug 18, 2022 9:06 am
- Forum: Chemistry forum
- Topic: Solubility product constant
- Replies: 8
- Views: 8421
Re: Solubility product constant
My answer to (a): Mg(OH)_2 + 2 e^- \leftrightarrow Mg + 2 OH^- E^{\circ} = -2.67 V oxidation Mg + 2 e^- \rightarrow Mg E^{\circ}= -2.36 V K = \frac{1}{K_{sp}} reduction log K = \frac{2 [-2.36 -(-2.67)]}{0.059} =10.51 K=\frac{1}{K_{sp}}= 3.2 \times 10^{10} K_{sp} = 3.1e-11 My answer to (b): Ag + 2 N...
- Wed Aug 17, 2022 10:25 pm
- Forum: Chemistry forum
- Topic: Dissolution of alkali metals in liquid ammonia
- Replies: 5
- Views: 4823
Re: Dissolution of alkali metals in liquid ammonia
So, (x + y) is a stoichiometry coefficient. Isn't it?
- Wed Aug 17, 2022 10:31 am
- Forum: Chemistry forum
- Topic: Dissolution of alkali metals in liquid ammonia
- Replies: 5
- Views: 4823
Dissolution of alkali metals in liquid ammonia
The alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature.
\( M + (x + y ) NH_3 \rightarrow [ M (NH_3)_x]^+ + [ e (NH_3)_y ]^-\)
What is x and y in the above equation?
How to interpret the above chemical equation?
\( M + (x + y ) NH_3 \rightarrow [ M (NH_3)_x]^+ + [ e (NH_3)_y ]^-\)
What is x and y in the above equation?
How to interpret the above chemical equation?
- Wed Aug 17, 2022 2:41 am
- Forum: Chemistry forum
- Topic: Na2O2 + CH3NO2 =?????
- Replies: 1
- Views: 1339
Na2O2 + CH3NO2 =?????
I know reactants and products of the following chemical reaction
\(3Na_2O_2 + CH_3NO_2 + 2H_2O \leftrightarrow HNO_3 + CO + 6NaOH \)
I want to know the product of the following reaction.
\(Na_2O_2 + CH_3NO_2 =???\)
Would any member of chemistry forum answer my question correctly?
\(3Na_2O_2 + CH_3NO_2 + 2H_2O \leftrightarrow HNO_3 + CO + 6NaOH \)
I want to know the product of the following reaction.
\(Na_2O_2 + CH_3NO_2 =???\)
Would any member of chemistry forum answer my question correctly?
- Wed May 25, 2022 11:49 pm
- Forum: Chemistry forum
- Topic: Solubility product constant
- Replies: 8
- Views: 8421
Re: Solubility product constant
I am working on a) and b).
(Deviation from the subject)What is PN? Would you explain?
(Deviation from the subject)What is PN? Would you explain?
- Wed May 25, 2022 9:43 pm
- Forum: Chemistry forum
- Topic: Solubility product constant
- Replies: 8
- Views: 8421
Solubility product constant
a)Calculate the solubility product constant for Mg(OH)2 Mg(OH)2 + 2 e⁻ → Mg + 2OH⁻ E° = -2.67 b) Calculate the formation constant for Ag(NH3)2⁺ How to answer both these questions? (Deviation from the subject) Why don't other moderators answer the questions asked by chemistry students, chemistry prof...
- Fri May 06, 2022 9:17 pm
- Forum: Chemistry forum
- Topic: Normality question
- Replies: 3
- Views: 12151
Re: Normality question
Though, your answer looks to me correct, my doubt is if 173.80 g is equivalent weight, how can we take it as a "Molar mass"? I think in this problem, we should assume (1equivalent acid/1 equivalent base) 20.07 mL x (1 L/1000 mL) x (0.1100 eq base/1 L) x (1 eq acid/1 eq base) x (173.8 g/1 e...
- Fri May 06, 2022 4:36 am
- Forum: Chemistry forum
- Topic: Normality question
- Replies: 3
- Views: 12151
Normality question
Find the percent acid (eq wt 173.80) if 20.07 mL of 0.1100 N base is required to neutralize 0.721 g of a sample.
What is the answer to this question?
What is the answer to this question?
- Sat Apr 16, 2022 5:24 am
- Forum: Chemistry forum
- Topic: Explanation required for 'AmmoniaConcentration' package in 'R'
- Replies: 0
- Views: 36034
Explanation required for 'AmmoniaConcentration' package in 'R'
Description of 'AmmoniaConcentration' package in 'R'. The total ammonia in aqueous solution is present in two chemical species: un-ionized ammonia, NH3 , and the ionized form, NH4 +. The NH3 species is the one more toxic for aquatic organisms, but current analytical methods do not permit measurement...