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Can Someone Solve This For Me On Chemistry? God Bless by TNORWAY: 9:24am On May 08, 2023 |
How long will it take for current of 745mA to deposit 8.56g of chromium into the cathode using a solution of CrCl3. It takes an electric current of 5.379A applied to a solution of MSO4 for 1hr to plate 7.0g of an unknown metal. identify the unknown metal. (M) |
Re: Can Someone Solve This For Me On Chemistry? God Bless by A4alpha: 12:28pm On May 08, 2023 |
No.1 How long will it take for current of 745mA to deposit 8.56g of chromium into the cathode using a solution of CrCl3.? To calculate the time required for a current of 745mA to deposit 8.56g of chromium onto the cathode, we will need to use Faraday's law of electrolysis. The formula is: n = (Q t) / (F M) where n is the amount of substance deposited (in moles), Q is the charge passed through the cathode (in coulombs), t is the time (in seconds), F is the Faraday constant (96485 C/mol), and M is the molar mass of the substance (in g/mol). The molar mass of chromium is 52 g/mol, so we can calculate the number of moles of chromium that need to be deposited: n = 8.56 g / 52 g/mol = 0.165 moles Next, we can calculate the charge needed to deposit this amount of chromium: Q = n F M = 0.165 mol 96485 C/mol 52 g/mol = 8.34 x 10 C Finally, we can calculate the time required to deposit this amount of chromium using a current of 745mA: t = (Q) / (I 3600) = (8.34 x 10 C) / (0.745 A 3600 s/h) = 312 s (or approximately 5 minutes and 12 seconds) Therefore, it would take approximately 5 minutes and 12 seconds for a current of 745mA to deposit 8.56g of chromium onto the cathode. 2. It takes an electric current of 5.379A applied to a solution of MSO4 for 1hr to plate 7.0g of an unknown metal. identify the unknown metal. (M)? Based on the given information, we can use Faraday's law to determine the unknown metal (M). First, we need to calculate the number of moles of electrons transferred during electrolysis using the formula: n = I x t / F where: n = number of moles of electrons transferred I = current (in Amperes) t = time (in seconds) F = Faraday's constant (96,485 C/mol) In this case, we need to convert the time from 1 hour to seconds (1 hour = 3600 seconds): n = 5.379 A x 3600 s / 96,485 C/mol n = 0.199 mol Next, we need to determine the molar mass of the unknown metal (M) using the formula: M = m / n where: M = molar mass (in g/mol) m = mass of metal plated (in grams) n = number of moles of electrons transferred (from above) In this case, we can solve for M: M = 7.0 g / 0.199 mol M = 35.2 g/mol Based on the molar mass, we can identify the unknown metal (M) as magnesium (Mg). Hope this help. 4 Likes |
Re: Can Someone Solve This For Me On Chemistry? God Bless by joinnow: 5:17pm On May 08, 2023 |
TNORWAY:From chatgpt For the first question, we can use Faraday's law of electrolysis to calculate the time required to deposit 8.56g of chromium at a current of 745mA. The formula for Faraday's law is: mass = (current x time x atomic mass) / (valence x Faraday's constant) where: current is the current in amperes (A) time is the time in seconds (s) atomic mass is the atomic mass of the metal in grams per mole (g/mol) valence is the valence of the metal ion in the electrolyte Faraday's constant is the charge of one mole of electrons, approximately equal to 96,485 coulombs per mole (C/mol) For chromium, the valence is +3, and the atomic mass is 52 g/mol. Plugging in the values, we get: 8.56g = (0.745A x t x 52g/mol) / (3 x 96485C/mol) Solving for t, we get: t = (8.56g x 3 x 96485C/mol) / (0.745A x 52g/mol) = 10,798 seconds ≈ 3 hours Therefore, it would take approximately 3 hours for a current of 745mA to deposit 8.56g of chromium into the cathode using a solution of CrCl3. For the second question, we can use the same formula to identify the unknown metal. The formula becomes: atomic mass = (current x time x Faraday's constant) / (mass x valence) Plugging in the values, we get: atomic mass = (5.379A x 3600s x 96485C/mol) / (7.0g x 1M) ≈ 55.8 g/mol To identify the metal, we need to find an element with an atomic mass close to 55.8 g/mol. The closest match is iron (Fe), whose atomic mass is 55.85 g/mol. Therefore, the unknown metal is iron (Fe). Therefore, the unknown metal is iron (Fe). 1 Like 1 Share |
Re: Can Someone Solve This For Me On Chemistry? God Bless by Omo9ja147(m): 6:05pm On May 08, 2023 |
joinnow:ogah..u good wolai |
Re: Can Someone Solve This For Me On Chemistry? God Bless by joinnow: 7:29pm On May 08, 2023 |
Omo9ja147:me don forget chemistry. I use chatgpt as stated from the beginning |
Re: Can Someone Solve This For Me On Chemistry? God Bless by Itulah(m): 8:58pm On May 08, 2023 |
TNORWAY: |
Re: Can Someone Solve This For Me On Chemistry? God Bless by Omo9ja147(m): 10:51pm On May 08, 2023 |
joinnow:u try cha.... Naso I Sabi chemistry too that year o .. Agriculture done carry my head now..mehn |
Re: Can Someone Solve This For Me On Chemistry? God Bless by ghettochild(m): 12:48am On May 09, 2023 |
joinnow:Please is chatgpt an app? |
Re: Can Someone Solve This For Me On Chemistry? God Bless by joinnow: 12:54am On May 09, 2023 |
ghettochild: is an AI that can be your personal assistant. it can help you in many many things. it can help you learn any skill provided you know how to give it prompt. to learn prompt go to learnprompt.org to make use o chatgpt go to chat.openai register and boom you have the world in the in hand. you can also use it to make money online. i interested let me know |
Re: Can Someone Solve This For Me On Chemistry? God Bless by falcon01: 5:59am On May 09, 2023 |
Am I the only person who chatgpt can't seem to give me correct answers to all my questions especially maths and physics. After solving some shit ill give another human and we arrive same answer but ChatGPT will write nonsense and give me and if I tell it that's its wrong it will Apologise and recalculate it in such a way that it will arrive at the answer that I did. |
Re: Can Someone Solve This For Me On Chemistry? God Bless by ghettochild(m): 6:11am On May 09, 2023 |
joinnow:I know what chatgpt is...just wanted to know if it's in the form of an app.. Anyways thanks |
Re: Can Someone Solve This For Me On Chemistry? God Bless by jaybee345(m): 8:21am On May 09, 2023 |
joinnow: I'm interested, please hook me up |
Re: Can Someone Solve This For Me On Chemistry? God Bless by joinnow: 8:28am On May 09, 2023 |
ghettochild:It has an app version and website version 1 Like |
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