English (deutsche Übersetzung unten!)
Today I would like to open up the next round of my #chemistrychallenge.
What could be more fun than activating your brain and at the same time having the opportunity to win up to 20 STEEM. :-)
Rules - what are the requirements to win the prize?
The winner is the first person to post the correct solutions (correct numbers and correct units; solution process required). The prize for the correct answers is 20 STEEM. I will send it to the winner immediately after the article is paid out (seven days after posting).
Should the various solutions be spotted by different users I will distribute the prize money of altogether 20 STEEM at my own discretion amongst the winners.
If nobody is able to determine the solutions before Thursday, February 1st, 2018 at 10 pm (German time!) the prize (20 STEEM) will be sent to the person with the best approach.
Special thanks to the SteemSTEM community! They support this challenge which means that they contribute half of the prize money (= 10 STEEM)! In my eyes this still growing science community adds a lot of value to our Steemit platform! If you are interested in 'STEM' (= Science, Technology, Engineering, Mathematics) you may check their blog and read their last project update.
Your Chemistry Challenge to Solve:
The mass of all worker bees within a beehive in summer is about 4 kg.
10 bees have an overall mass of 1 g.
Within 20 days one single bee collects enough nectar to produce 9 g of honey (about one teaspoon).
Let's assume that the honey contains 80 % (mass percentage) sugar from which 37.5 % are glucose (C6H12O6). The molar mass of glucose is M(C6H12O6) = 180 g/mol.
Questions:
a)
Let's assume that all worker bees of this beehive would collect nectar simultaneously.
- How many minutes would they need to collect enough nectar to produce exactly 500 g honey?
- How many mols and how many molecules of glucose would that amount of honey contain?
b)
The 500 g honey fit into a cylindrical honey jar.
What is the density of the honey? |
c)
Now let's assume that you eat exactly one-tenth of the honey in this honey jar to be fit enough to solve the last part of the challenge. :)
In your metabolism the glucose contained in the eaten honey is now oxidized by means of oxygen to CO2 and H2O (process of cellular respiration).
Which amount of energy ΔHR will be released from the eaten glucose during this degradation process?
Let's assume that under the given conditions (pressure, temperature) the following molar enthalpies of formation would apply:
ΔHf(H2O) = - 285.9 kJ/mol.
ΔHf(CO2) = - 393.5 kJ/mol.
ΔHf(C6H12O6) = - 1268 kJ/mol.
Deutsch
Heute möchte ich die nächste Runde meiner #chemistrychallenge eröffnen.
Was könnte mehr Spaß bereiten, als mal wieder sein Gehirn zu aktivieren und dabei zugleich die Gelegenheit zu ergreifen, bis zu 20 STEEM zu gewinnen. Also macht euch an die Arbeit! :-)
Regeln - welche Bedingungen müssen erfüllt sein, um den Preis zu gewinnen?
Gewinner ist derjenige, der zuerst die korrekten Lösungen (korrekte Zahlen und korrekte Einheiten; Lösungsweg erforderlich) nennt. Der Preis für die korrekten Antworten beträgt insgesamt 20 STEEM. Ich werde ihn dem Gewinner direkt nach der Vollendung der Auszahlungsperiode dieses Artikels (also sieben Tage nach seiner Veröffentlichung) gutschreiben.
Sollten die verschiedenen Lösungen von verschiedenen Usern gefunden werden, verteile ich das Preisgeld von insgesamt 20 STEEM nach eigenem Ermessen unter den Gewinnern.
Solle niemand die Lösungen vor Donnerstag, den 01.02.2018 um 22:00 Uhr (deutsche Zeit!) herausfinden, wird der Preis (20 STEEM) demjenigen gutgeschrieben, der den besten Lösungsansatz vorzuweisen hatte.
Mein besonderer Dank gilt der SteemSTEM-Community, die diese Challenge unterstüzt, indem sie die Hälfte des Preisgeldes (= 10 STEEM) beisteuert! In meinen Augen ist diese noch wachsende Wissenschafts-Community ein wertvoller Bestandteil der Steemit-Plattform! Wenn du dich für 'STEM' (= Science, Technology, Engineering, Mathematics) interessierst, empfehle ich dir, ihren Blog anzusehen und das letzte Projekt-Update zu lesen.
Hier nun die zu lösende 'Chemistry Challenge':
Die Masse aller Arbeitsbienen eines Bienenvolks im Sommer betrage 4 kg.
10 Bienen haben eine Masse von zusammen 1 g.
Innerhalb von 20 Tagen sammelt eine einzelne Biene genug Nektar, um davon 9 g Honig (ungefähr den Inhalt eines Teelöffels) produzieren zu können.
Angenommen, der Honig enthalte 80 % (Massenanteil) Zucker, wovon 37,5 % Glucose (C6H12O6) seien. Die molare Masse von Glucose beträgt M(C6H12O6) = 180 g/mol.
Fragen:
a)
Nehmen wir an, alle Arbeitsbienen des Bienenvolks würden gleichzeitig Nektar sammeln.
- Wie viele Minuten würden sie benötigen, um genug Nektar zu sammeln, um davon genau 500 g Honig zu produzieren?
- Wie viel mol und wie viele Moleküle Glucose wären in dieser Honigmenge enthalten?
b)
Die 500 g Honig passen in ein zylinderförmiges Honigglas.
Was ist die Dichte des Honigs? |
c)
Lasst uns nun annehmen, du würdest genau ein Zehntel des in diesem Glas enthaltenen Honigs essen, um fit genug für den letzten Teil dieser Challenge zu sein. :)
In deinem Stoffwechsel wird nun die im verzehrten Honig enthaltene Glucose mit Hilfe von Sauerstoff zu CO2 und H2O oxidiert (Prozess der aeroben Atmung).
Welche Energiemenge ΔHR wird bei diesem Abbauprozess der gegessenen Glucose freigesetzt?
Lasst uns weiterhin annehmen, unter den gegebenen Bedingungen (Druck, Temperatur) herrschten folgende molare Bildungsenthalpien:
ΔHf(H2O) = - 285,9 kJ/mol.
ΔHf(CO2) = - 393,5 kJ/mol.
ΔHf(C6H12O6) = - 1268 kJ/mol.
Hey Jaki!
Offen und ehrlich gesagt wäre ich bei dieser Aufgabe komplett verloren und hätte da null Chance. Zumal ich auch noch viel zu spät dran wäre :)
Allerdings verdient alleine deine Arbeit solch eine Aufgabe zu stellen und solch einen Post zu schreiben die volle Unterstützung 😁
100% UPVOTED 👍
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Let me try to give it a shot:
Q1
1 bee implies 0.1 gram; 4 kg bee implies 40000 bees
honey production rate = 9/20 gram/bee/day, implies 0.08 min per gram of honey.
500 gram honey = 150 gram glucose
Q2
Density = mass / vol = 500 /(Pi(4)27.958)
Q3
Enthalpy of reaction:
C6H12O6 + 6 O2 --> 6CO2 + 6 H2O
by completing the Hess cycle,
The required enthalpy = (-393.5)(6) + (-285.9)(6) - (-1268)
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Opus I forgot that was 1/10 of the honey. So I edit add a few line here:
Consumed honey = 50 gram = 15 gram glucose = 0.0833333 mole =
so ΔHR = -234 kJ
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Right! So @jedigeiss 'stole' you this part of the solution because he was faster! :)
Well done nevertheless - hope you enjoyed the competition! (I will send you 3 STEEM.)
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Haha !! That was really fun!!! This is probably the first time for me to work on a Hess cycle in the recent five years. Really took me some time back to the definition and make sure I didnt remember things wrongly XDD
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A 1:
6250 min40 minA 2: 0.8333.. mol and 4.99821023 molcules
B: 0.3125g/cm*3
C: -48.85 kJ/mol
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B: 500g/400cm ** 3 = 1.25 g/cm ** 3
(4**2 * 3.141 * 7. 958)
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That looks better now ... :)
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I suggest you to check the exact number of the molecules (part a)) again.
Also part c) is still wrong ...
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Tonight.
For the chemistry stuff I have to get back to Wikipedia and reread some parts.
I guess someone is already picking up my mistakes :))
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Part of the price money is already yours ...
Part c) is most difficult maybe ...
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Nice that you tried it, but most of it (not everything) is still wrong ... (I didn't give you more details for now not to give anybody any early hint ...)
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That's the best I could do on my mobile phone :)
Will try later if no one answered
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So far without any doubt your answers are the best ones (not to say: the only ones). :)
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Can you check my answer in steemd?
Because the markup here is destroying my * 10**23
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You can write it by using <sup>text</sup>
1023
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Good that you didn't delete completely but striked out the old solution so that I can follow your solving process better. However, there are still some wrong answers ... :)
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That bloody
AvocadoAVOGADRO .. Why isn't it a well rounded number like Pi!Oh.. I see.. :)
EDIT: Avogadro and avocado are spellchecker mistakes on my phone.. Sorry.. Lol:)
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Top and bottom of the cylinder are circle areas ... That part shouldn't be toooo difficult ...
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Lol, that was funny indeed! :)
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hi @jaki01, crap that i saw this now i need to try to solve it.
for a its 40 min, thats correct.
a2,0,833 periode is also right, i get 5,018 * 10^23 molecules
b,die Dichte beträgt 1,24 g /cm3
c,49,04 kJ
i dont know, i was never brilliant in chemistry ..
;D
Kind regards
Jan
new answer
well since C6H12O6 + 6 O2 --> 6CO2 + 6 H2O is correct and 2808,4kJ/mol as well
you only need to see that you are taking a tenth of 0,83 period of mol meaning 0,083 period
this results in 234 kj
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Well done! However, @isnochys found some of the solutions faster! :)
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Hi @jaki01. Here my answers:
a) t=40min, 0.833 mols and 5,016x1023 molecules
b)ρ=1,250 g/cm3
c)ΔHr=-2808,4 KJ/mol
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is it late?
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Part c) is incorrect. The correct answer is -233KJ as @jedigeis said!
I apologize.
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No need to apologize! You did well, but the some other Steemians were faster this time! :)
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Chemistry is not my strong, I prefer mechanics, but I like very much thermodynamics , it's ironic, right?
Thanks for this challenge @jaki01, it was very interesting.
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Really nice to hear that you liked it ...
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A. 1. 45 mints..
2. 0.9457. mol and 4.12654 molcules
B. 0.3125g/cm3
C. 24.5kj/mol
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Thanks for taking part as well, but your solutions are not yet correct. :)
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Ohhh...
My chemistry knowledge is not good...but....I can try again?...
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Sure!
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A: (1) 40 minutes. (2) 2.78 mol, 1.67*10^24 molecules
B: 1249.96 kg/m3
C: -780.11 KJ
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A
1: for time
no of bees = 40,000
in a 20 day period the weight of honey produced by them = 360000 g
using the direct relation between time and wt, we get 1/36 day = 24/36 hr = 40 min
2: for no of moles
mass percentage of glucose= 80%of37.5%of500= 150g of glucose
no of mols= 150/180= 0.833 mol
no of molecules= 150/180* Avogadro's number= 5.091*10^23 molecules
B
volume of cylinder = pir^2h where r=0.04m
density = mass/volume = 1249.96 kg/m3
C
From Hess's law
Energy released when 1 mol of glucose is burned is -2808.4 KJ
since, 1/10th of honey is equivalent to 15 g of glucose, using the simple unitary method we get -234.033 KJ
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Apart from the number of molecules that looks really good! However, some other Steemians were faster this time! :)
I hope you enjoyed the competition nevertheless!
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Yeah i did. Thank you for encouraging this kind of posts. TBH, i dont know how to use the power function here, so just used raised to the '^'
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You can write it by using <sup>text</sup>
For example: 1023
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Thank you :). And about the number of molecules, i was contemplating till now and realized what you meant. But here in our country, we're taught that the base of Avogadro's no is 6.023 and not 6.022. This was in high school and it's still embedded in me. And again Thank you and will keep an eye out for your contests. CHEERS!
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A:(2) 0.833 mol , 5.091*10^23 molecules
C: -233.7(-234.033, my bad) KJ, bad habit of skipping lines.
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haha know my 2 is wrong. I'll redo it. Didn't see the mass % part. Also C. 1/10 th of the honey not glucose.
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@jaki01 : since i did in my notebook, do i need to post the photos?
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Ich muss mich leider aus dem Wettbewerb verabschieden, ohne überhaupt teilgenommen zu haben.
Chemie hatte ich nur bis zur Mittelstufe und Mathe ist auch heute noch das Fach, durch das ich mich im Studium kämpfen muss haha
Vielleicht hab ich beim nächsten bessere Chancen :P
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Wie? Was? Eine Stunde und noch keine korrekte Antwort? Wo sind unsere Wissenschaftler heute? da habe ich ja direkt eine Chance. Da ich auf dem Steem als multible Persönlichkeit agiere, fangen wir mal an. Der Frosch, Afro–G, das Orakel, der Witnessblower und ich. Zusammen werden wir das doch schaffen!
4000g x 10 = 40000 Bienen.
In 20 Tagen, also 28800 Minuten sammeln 40000 Bienen 9g x 40000= 360000g Honig!
360000/100 = 3600 x 8 = 28800g Zucker, mit 10800g Glukoseanteil (37,5%) was 60 mol sind (molekulare Masse).
1 mol sind 6,022 · 1023 Teilchen, Atome
Sie sammeln innerhalb von 28800 Minuten 360000g Honig.
360000/28800 = 12,5 g/min
500/12,5 = 40
Antwort a) 40 Minuten. 60 mol und
Ich habe an dem Punkt aufgegeben:
180,16 g·mol−1.
Habe noch meinen Freund angerufen, der war Mathematiker und der meinte ich soll durch das, was davor steht dividieren. Hoch minus Eins. Die Wurzel aus Eins ist es noch nicht. Aber so was Ähnliches.
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Sehr gut, also, die Minutenzahl ist schon mal richtig!
Als Molmasse der Glucose kannst du den von mir gegebenen Wert 180 g/mol nehmen.
Allerdings musst du erst die Honigmasse, 500 g, mit 0,8 und dann mit 0,375 multiplizieren, um die Glucosemasse in 500 g Honig zu berechnen. Danach gilt: Die Anzahl der Mole ist Masse dividiert durch Molmasse, also n = m/M = 150 g / 180 g/mol = 5/6 mol.
Allerdings ist das schon gelöst worden, siehe oben!
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Du weißt doch, lieber @Jaki01: Ich bin seit Monaten viel zu langsam. Beim Schach ist auch schon immer alles gelöst. Trotzdem hat es immer Riesenspaß gemacht.
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Vor etwas mehr als 25 Jahren hätte ich eventuell eine Chance gehabt, Dir eine Lösung zu präsentieren.
Ich hatte LK Chemie (14 Punkte - ich war damals fleißig; und organische Chemie hat mir Spaß gemacht) und sogar ein Semester Chemie an der Uni studiert.
Das Wissen von damals ist allerdings bei mir völlig ausradiert!
Völlig...!
Ich habe an der Uni ganz schnell gemerkt, daß Chemie zwar eine tolle Sache ist - aber nicht für mich! ;-)
Und alleine wenn ich diese Aufgabe durchlese, weiß ich warum ich schnell das Weite gesucht habe... :-) :-) :-)
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:-)
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Ich bin ehrlich mein lieber jaki...... bereits beim lesen der Fragen, hatte sich mein klägliches Gehirn abgeschalteten......ABER.... immerhin habe ich es gelesen... :-)
PS. Die Lösung lautet Schwarzbier******
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Nicht nur mathematische Fähigkeiten sind, sondern auch Humor ist eine positive Eigenschaft! :)
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wo ich natürlich völlig bei dir bin...... :-)
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Da muss ich absolut passen lieber @jaki01 :-) Das war noch nie meine Stärke. Ich bin gespannt wer das raus bekommt. Sollte bestimmt machbar sein. Mein Lehrer hatte damals zu mir gesagt vor der ganzen Klasse dass ich ein "Matheschenie " sei, bei solchen Aufgaben. also, du weist nun wie "gut" ich in sowas bin. Bei so etwas bin ich kläglich überfordert.. Freue mich wer es heraus bekommt! Good Luck all ;-)
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Das Schöne ist, dass verschiedene Menschen verschiedene Stärken und Vorlieben haben ... :)
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I was a chemistry student and I have to look into it through books again to solve it out.
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Ich bin zwar beim ersten Satz der Rechnung schon ausgestiegen, aber dafür wollte meine Mutter direkt wissen, was es hier mit Bienen zu erzählen gäbe :D
Viel Glück dem Gewinner und ganz liebe Grüsse aus der Schweiz <3
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Es ist großartig zu sehen, dass diese schwierige Aufgabe von so vielen Leuten bearbeitet wurde. Nun forderst du also die Community nicht nur mit Schach, sondern auch noch mit STEM Tüfteleien heraus. Einfach nur Wahnsinn!
Ein Bild sagt mehr als 1000 Worte!
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Danke - freut mich, dass dir die Aufgabe gefällt (ich habe mir einige Mühe gegeben, der Mathematik und Chemie einen möglichst interessanten Rahmen zu verpassen)! :)
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I was poor in chemistry in college and school days but I like this type of contest and this sort of contest really uplift the standard of steemit.
Thank you...steem on and stay blissful....
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Great puzzle. Too bad I read it a bit too late to contribute, but definitely a fun set of questions.
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I am glad to read that you like it! :)
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Ich lass die chemestrychallenge mal ausfallen, bin aber bei der chess challenge immer dabei. Die ist eher was für mich :D
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Sehr coole Aktion! Wenn ich morgen mehr Zeit habe versuch ich mich mal dran
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My plan is not to look at the comments section of this post until after the payout and work on this in my off time. Aversion to even simple math tends to be the reason some people become non-tax lawyers in the first place, so I have low expectations for myself :)
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Just take your time and enjoy the riddle. :)
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To be honest I am very jaded in chemistry Sir @jaki01, the start of high school. I usually always take the time to find the answers of your questions through the postings. But this time I surrender Sir, I am unable to answer it. :)
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You got a 10.77% upvote from @mercurybot courtesy of @b-s!
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Great challenge.
Many smart steemians that already answered correctly.
How do we find real pure honey in this modern world of deceptions and fake honey?
Do you know any easy at home method to verify if I bought real honey or not?
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The problem is that there are many ways to manipulate honey, that means to be really sure you would need to make not only one but a few tests ...
Some of the possible tests you can find here.
The best method to be sure though, is to buy it directly from my father. :-)
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Thanks, those tests are pretty simple for home use.
But to be sure I best find your father and buy honey from him ;-D
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oh such a great idea Mr jaki :) congratulation to winners
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Very nice post, @joki.Thanks to sharing it with us.And very difficult question chemistrychallenge.Keep it up,buddy.
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Very challenging question it,@joki01.But i dont know its answer.Very good post.Keep it up.
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Hello @jaki01,
Extraordinary good question & a contest. Incredible contest idea & a tag.
~@mywhale
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wow chemistry challenge great but my chemistry is week😐😐
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Which week? This week or next week? :)
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Thank you for this logical analysis.
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great to share chemistrychallenge
thank you
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Chemistry is a very difficult subject. I understand chemistry very little. It is understood that you understand a lot about chemistry.
keep it up & thank you.
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quite difficult friends,
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@jaki01,
I assure I don't know the answer for this question! But I hope my sister knows it :D Hope she will try this contest!
Cheers~
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Very challengable question & you share a nice prize for it
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Good question & Excellent contest. Brilliant article~
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@jaki01 - Sir it's a good contest idea... Chemistry is not a good subject of mine... But Sir, I think this is the 1st contest of Chemistry runs on a blockchain... Nice idea Sir...
+W+
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this post is very knowledgeable post.
thanks for sharing.
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excellent contest idea & looking for answer to this question. impressive post @jaki01
@resteemia
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my brain not working ...its too hard ....
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I do not understand your question. .. ??
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i got to learn a lot from it thanks for sharing and congratulations winners
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wow ... it's hard to be the fastest in this contest ... incredible for you fastest, and for you @jaki01, you have a contest that I think is very special ... amazing one for you @jaki01
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Nice post, congrats to winner of this contest.
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i thought i could win this but winners already won it hahah :D
congratulations to them
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i need to work hard on these challenges thanks for sharing bro
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No knowledge of chemistry . Thanks for post
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i got really late in this challenge :(
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i literally all the answers in the first comment hahh otherwise i would start solving them lol
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This is great post and useful
Thanks @jaki01
Have a nice day
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After the chess quiz, now there is another new quiz.. @jaki01
It seems I should damn well learn quiz, brother need a process that is quite time consuming to answer his
Because I am less steeped in this field ... @jaki01
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congratulations to all the winners they are incredible :) and the host too ;)
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That is a interesting post .thanks for sharing
For your post propagation.
Upvote/Resteem
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You got nominated in my latest article TOP 5 trending Science Accounts you should follow today!
Guess who is the winner!
Feel free to check it out:
https://steemit.com/science/@bestmemes/top-5-trending-science-accounts-you-should-follow-today
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Excellent article ,good question & a great contest.
Great work sir
Resteem
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Great post. it is very valuable information
Resteem service
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This post is not idea. thanks for sharing
@jaki01
Upvote Resteem
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Greetings @jaki01. According to my knowledge the answer is
A. 40 minutes, 0.884 molecules and 5.02 x 10^23 molecules
B. Ρ = 1.250 g/cm3
C. ΔH1 = -2800 KJ
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Well done, most of you answers are correct. The number of moles is 0.833 mol.
Part c) is wrong, as you didn't consider, that the eaten honey is only one-tenth of 500 g which contains far less than 1 mol glucose ...
Check the solution above!
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Ok, Thanks for the correction, sorry my chemistry is still very lacking
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thanks for sharing this challenge i tried solving it but failed
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wow this is a cool challenge everyone could participate :)
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thanks for sharing this contest .... congrats to the winners
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Excellent post and helpfully
100% like and resteem
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its was a nice challenge for me need to learn
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You're take this great chemistrychallenge.
Thanks for share.
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All the best for take this great chemistrychallenge.
god bless you and I think,You will be the winner.
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I hated chemistry and really now I regret because I lost the oppurtunity of earning 20 steem but really reading so many answers I learned a lot...... @jaki01 you have really got skill of engaging people in your post even if it is a chess blog or this everybody does not have this skill I learn a lot from you bro keep it up :)
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I think
A. 1. 45 mints..
B. 0.3125g/cm3
C. 24.5kj/mol
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Exactly the same results like @dini1 (even the same spelling mistake: 'molcules'), are you sure?
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not sure but i think @dini1 correct.
please details
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Own solutions please ...
The winners are announced by the way - see above.
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i find your challenge today... 😖 my bad luck... 😭😭😭
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Congratulations @jaki01! You have completed some achievement on Steemit and have been rewarded with new badge(s) :
Award for the number of upvotes
Click on any badge to view your own Board of Honor on SteemitBoard.
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thanks for your information @jaki01
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i dont no know of chemistry. thanks for this post @jaki01
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thanks for your post dear friend.
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Find ich ja sehr geil sowas. Leider bin ich viel zu spät dran, das nächste mal seh ich deinen post hoffentlich füher. Trotzdem Danke.
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@jaki01 Please note STEEM is #25 in coinmarketcap.com and have big peace of market over $1.5 billion." - it's piece, not peace. Thanks for trying, but I simply don't think they will consider it because of the points you're giving them.
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Thank up information llike you
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A 1: 6250 min 40 min
A 2: 0.8333.. mol and 4.99821023 molcules
B: 0.3125g/cm*3
C: -48.85 kJ/mol
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Please, DON'T copy the solution from other participants (@isnochys in this case)! That's plagiarism!
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Very exiting quiz. I'm interested in chemestry, but I feel that I have not enough knowledges and skills to complete this task. Danke schön!
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The information is very interesting I like it @ jaki01
You are a great friend
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Brilliant writing @jaki01, good job
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nice plz sir one vote plz sir....
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You beg for upvotes even if you didn't upvote my article? Come on ... :)
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The information is very interesting @jaki01
you are indeed terrific..
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Awesome challenge for dynamic minds, the formulars stated are great, i'll my geneus chemistry student, i'm sure he'll attempt this
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i think you create a great post...i learn many thing from your post..and i impressed to see it...thanks for shear this useful thing...
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Ilmu kimia yang sangat bagus, # @jaki01
Saya sangat menyukai blog anda
Kalo ada waktu silahkan kunjungi blog saya
https://steemit.com/latest/@cutminajn/xpander-specifications-fcd10503a8c0b
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What about upvoting mine first? :)
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