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Natriumhydrgencabonat - Physics

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Natriumhydrogencarbonats omdannelse
Af Nynne Viktoria Sass Larsen, 1.d

Formål

Formålet med opgaven er at finde hvilket af reaktionsskemaerne der er den rigtige kemiske reaktion ud fra en teoretisk og en praktisk del. Vi skal hermed bestemme hvilken omdannelse der finder sted, når stoffet NaHCO3 bliver ophedet.

Teori

NaHCO3 bedre kendt som natron, bruges som et hævemiddel i bl.a. kager. Stoffet består af et salt, altså en ionforbindelse, og kulsyre. Årsagen til at dette bruges som hævemiddel, grundes i at stoffet under ophedning afgiver gasser, så så får eks. en kage til at hæve.

For at bestemme hvilken reaktion der forekommer fra dette forsøg på et teoretisk plan, skal vi bruge en række formler, for at bestemme de forskellige vægte mm. De formler jeg gør brug af i denne opgave lyder følgene;

n betyder indenfor fysikkens verden, stofmængde. Det er et udtryk for et antal af formelenheder og dets SL-enhed er i mol. En formelenhed er et stof ens med den mængde af dette stof, som angives i dets sumformel. I reaktionsskemaer angiver vores koefficienter forholdet mellem stoffets mængder. Eks;

NaHCO3(s) Na2O(s) + H2O(g) + CO2(g) = 2NaHCO3(s) Na2O(s) + H2O(g) + 2CO2(g)

I dette afstemte reaktionsskema, kan vi se, at NaHCO3 fordobles for at afstemme skemaet korrekt. For at beregne n benytter vi forholdet mellem stoffets molarmasse(M) og masse(m). Formlen lyder således;

n=mM

m er et udtryk for massen af et stof. Massen er et begreb for mængden af stof. I det periodiske system, kan massen af et atom aflæses. Til dette ved vi, at massen af et atom afhænger af protoner og neutroner. På den måde kender vi vægten på grundstoffet/atomet. Vi måler massen på vægt og dets SL-enhed er i gram(g) eller kilogram(kg). Massen kan findes ved at kende stofmængden(n) og molarmassen(M). Denne formel lyder således; m=n*M I vores forsøg har vi dog kendt massen af stoffet på forhånd, og det har ikke været nødvendigt at beregne stoffets masse.

M betyder molarmasse. Molarmasse er en betegnelse for de sammenlagte molaremasser i et givent stof, ud fra dets grundstoffer. Enheden g pr. Mol, står for antallet af gram pr. Mol vi har. For at beregne den molaremasse, sammenlægges hvert grundstof der indgår i stoffets molarmasser til et samlet resultat. Dette resultat er stoffets molarmasse. Molarmassen kan dog også beregnes, hvis massen og stofmængden af et stof kendes;

M=n*m

Molarmasse, stofmængde og masse kan som overstående ses, have forhold imellem sig, som vi kan bruge. Molarmassen er ensbetydende med masse pr. Mol, og med det for vi følgende formler;

n=mM og M=n*m og m=n*M

Man kan formulere forholdet således; massen af et stof er ens produktet af stofmængden og stoffets volumen. Understående billede viser også forholdet mellem massen, stofmængden og molarmassen.

En kemisk reaktion kan skrives som et reaktionsskema. Et reaktionsskema afstemmes ved at have en ens mængde af grundstoffer, på hver sin side af reaktionspilen. Stofferne der reagerer i dette skema kaldes reaktanter og de dannede stoffer kaldes produkter.

Reaktanter produkter

På venstre side af reaktionspilen står den uafstemte del af reaktionsskemaet, og på højre side findes det afstemte, altså vores produkt. Et reaktionsskema er afstemt, når der er lige mange atomer af samme slags, på hver sin side af reaktionspilen og når den elektriske ladning er ens på begge sider. Afstemningen gøres ved angive koefficienter foran vores stoffer. Tilstandsformer angives også i reaktionsskemaerne for de enkelte molekylers tilstand. S er fast form, l er flydende, g er gasform og aq er hvis stoffet er opløst i vand.

Afstemte reaktionsskemaer;

1. NaHCO3(s) Na2O(s) + H2O(g) + CO2(g) = 2NaHCO3(s) Na2O(s) + H2O(g) + 2CO2(g) 2. NaHCO3(s) NaOH(s) + CO2(g) = ingen ændring 3. NaHCO3(s) Na2CO3 + H2O + CO2 = 2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

Materialer

* vægt * bunsenbrænder * porcelænsdigel * trekant * trefod * digeltang * natriumhydrogencarbonat

Udførelse

Vi tog først den tomme digel og satte den i trekanten på trefoden og opvarmede den i et minuts tid. Hvis der skulle være eventuelle rester af stof ville det forsvinde efter denne opvarmning. Efter diglen var afkølet vejede vi den. Derefter fyldte vi diglen med natriumhydrogencarbonat ca. 5g, og satte igen diglen på trekanten. Vi opvarmede så diglen i 10 minutter, og da den var afkølet afvejede vi igen diglen.

Resultater og databehandling

Resultater | Før ophedning | Efter ophedning | Masse af digel | Masse af digel og NaHCO3 | m(NaHCO3) | Masse af digel og stof X | mmålt(X) | 14,564g | 19,571g | 5,007g | 17,719g | 3,155g |

For at finde m(NaHCO3) trækker jeg massen af diglen fra massen af digel og NaHCO3. For at finde mmålt(x) trækker jeg massen af diglen fra massen af digel og stof efter ophedning.

Natriumhydrogencarbonat | M(NaHCO3) | n(NaHCO3) | 83,99 g/mol | 0,06 mol |

Vi har taget alle grundstoffernes molarmasse og lagt dem sammen så M(NaHCO3) = 83,99 g/mol. M(NaHCO3) =22,99+1+12+16*3g/mol=83,99 g/mol
For at finde stofmængden bruger jeg formlen n=mM . Jeg indsætter de fundne værdier: nNaCHO3=5,007g83,99g/mol=0,06 mol

Mulige produkter | Fast reaktions produkt X | Reaktion 1 X er Na2O | Reaktion 2X er NaOH | Reaktion 3X er Na2CO3 | mber(X) | 1,86g | 2,40g | 3,18g | M(X) | 61,98 g/mol | 39,99 g/mol | 105,99 g/mol | n(X) | 0,03 mol | 0,06 mol | 0,03 mol |

Nu skal jeg finde den teoretiske masse af de forskellige stoffer(X) i reaktionsskemaerne, reaktion 1, 2 og 3. N(stofmængden) af Na2O er fundet ved at aflæse reaktionsskemakoefficienterne.
Så for reaktion 1 er forholdet mellem NaHCO3 og Na2O 2:1, dvs n(Na2O)=0,03. For reaktion 2 er forholdet 1:1, dvs n(NaOH)=0,06. For reaktion 3 er forholdet 2:1, dvs. n(Na2CO3)=0,03.
For at finde molarmassen i reaktion 1 M(Na2O)=22,99*2+16,00g/mol=61,98 g/mol. For at finde molarmassen i reaktion 2 MNaOH=22,99+1,00+16,00g/mol=39,99 g/mol. For at finde molarmassen i reaktion 3 M Na2CO3=22,99*2+12,01+16*3g/mol=105,99 g/mol.
For at finde massen af X for hver af reaktionerne benytter vi os af formlen m=n*M.
For at finde massen i reaktion 1, mNa2O=61,98gmol*0,03 mol=1,86g. For at finde massen i reaktion 2, mNaOH=39,99gmol*0,06mol=2,40g. For at finde massen i reaktion 3, mNa2CO3=105,99gmol*0,03 mol=3,18g.

Fejlkilder og diskussion

Nu har vi udregnet både den teoretiske del og udført den praktiske del af forsøget. Vi kan nu se, at selvom vi er kommet frem til et svar til hvilket reaktionsskema der passer overens, så er der en mindre afvigelse. Vores teoretiske svar er på 3,18g og vores praktiske svar er 3,16g. Den procentvise afvigelse lyder således;
Procentvis afvigelse:3,16-3,183,18*100=-0,62%

Vi har en procentvis afvigelse på 0,62% fra teoretisk svar til praktisk svar. Selvom 0,62% ikke er en relativt stor afvigelse, så betyder det stadig, at der er begået en fejl i vores forsøg. Mulige fejlkilder til dette kunne eksempelvis være forkerte beregninger, forkert afmåling eller at vores ophedning af stoffet ikke har stemt overens med tid eller den tilførte varme. Da jeg har billeder af vægten på vores forsøg, og jeg ikke husker at tabe noget, så vil jeg udelukke at dette kan være årsagen til vores forskel fra teori til praksis. Jeg vil heller ikke mene, at mine beregninger er forkerte, så den fejlkilde vil jeg også udelukke.
Det er meget muligt, at der er opstået noget under vores forsøg, der er årsagen til denne forskel. Men da vores teoretiske svar er større en det praktiske, vil jeg udelukke muligheden, at vi ikke afbrændte vores stof længe nok. Dette ville have resulteret i, at vores omdannelse af NaHCO3 ikke ville være færdiggjort, og vores praktiske svar ville dermed have været større end det teoretiske. Ud fra mine informationer kan jeg ikke afgøre, hvad årsagen til forskellen er .

Konklusion

Ud fra forsøgets praktiske og teoretiske del, kan jeg nu konkludere at reaktion 3 er den omdannelse der finder sted under ophednings processen. Det har lykkedes os at finde den rette omdannelse ved at sammenligne teori og praktisk udførelse.

--------------------------------------------
[ 1 ]. Alle formeler brugt i databehandling kan findes i teorien.

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...Final Paper Albert Einstein and Isaac Newton were both very influential figures concerning science. They both discovered ground breaking things in the physics world. Albert Einstein was a German-born theoretical physicist who developed the general theory of relativity. This is one of the biggest parts of physics alongside with quantum mechanics. Sir Isaac Newton was an English physicist and mathematician who are commonly referred to as one of the most influential scientists of all time as well as a key figure in the scientific revolution. Newton formulated the laws of motion and the universal gravitation that dominated scientists’ view of the physical universe for over the next three centuries. He also has demonstrated that the motion of objects on the Earth and that the celestial bodies could be described by the same principles. When he was deriving Kepler’s laws of planetary motion from his mathematical description of gravity, Newton removed any of the people’s last doubts about the validity of the model of the cosmos that was heliocentric. Near the start of Albert Einstein’s career he was beginning to think that Newtonian mechanics was no longer enough to reconcile the laws of classical mechanics with the laws of the electromagnetic field. While he was doing this it led him to his special theory of relativity. Thus he realized that the principle of relativity could also be extended to the gravitational fields, and this sparked his subsequent theory of gravitation in 1916...

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