Can anyone help me with these chemistry problems?
4 mole of Aluminum requires 801 kcal 1.07 mole of Aluminum requires x 4/1.07 = 801/x 4x=1.07 * 801 x = 214.3 kcal Problem Two ======== 1 mole of aluminum = 27 grams x mole of aluminum = 10.6 grams 1/x = 27/10.6 x = 0.393 moles of aluminum 4 moles of aluminum requires 801 kcal 0.393 requires x 4/0.393 = 801/x x = 78.6 kcal
Need help with organic chemistry. I don't get what the questions below is asking me to do, can anyone explain it???
these two questions have to do with "fischer projections" see these links https://en.wikipedia.org/wiki/Conformati... https://en.wikipedia.org/wiki/Cyclohexan... https://en.wikipedia.org/wiki/Fischer_pr... for the left, note both Br's are above the ring. now draw both boat and chair confirmations (see the 2nd link). now determine which is lower energy (1st link). for the right, convert that to the fischer projection. note whether the Cl's are coming out of or into the page. It's tricky. You need to pay attention to that conformation. make sure you draw out and understand where the H's are.
Chemistry help....can anyone explain the steps in solving this problem?
All you are asking basic questions in gaseous state. Here both pressure and temperature are changing, so assume ideal behaviour and combine both Boyle's law and Charle's law. common formula is P1V1 / P2V2 = n1 R T1 /n2 R T2 this is the common formula you can apply to any ideally behaving gasses here 'n' is constant the means n1=n2. R is always constant. cancell the equal terms then P1V1 / P2V2 = T1 / T2. initial conditions are STP ie....... P1 = 1atm T1 = 0 C = 273 k V1 = 20 lit final conditions are P2 = ? T2 = 50 c = 50+273 k = 323 k V2 = 20 lit substitute and find P2
Empirical Formula Help?
http://dbhs.wvusd.k12.ca.us/webdocs/Mole/EmpiricalFormula.html the sight explains it all though, some additional help may be needed: mass to moles step: find the molecular mass on the periodic table, for instance carbon would be 12.01, then divide the amount you have by it. (1.75/12.01=0.145711)(bromine: 46.75/79.9=0.585106) divide by small: the smallest would be 0.145711(carbon) so you divide each of the moles by that. you should get 1 for carbon and 4.o155238 for Br. since Br is less than any significant digit(rounding wise) you leave it, so it wouldl be Br4C. Note: if you get a number that is .1 away from a number, you go with the newer number. (ex. 2.9-->3) hoped i helped better than the first guy.
Electron Configuration help! Can anyone explain how to figure this out?
a) The first level ( or the first row of the periodict table) has one orbital s that can hold 2 electrons: 1 s2 b) The second level has 2 kinds of orbitals : s and p. The total electrons : 8 , two of these into the 2s orbital, six into the 2p. c). The third level has 3 kinds of orbitals : s, p , and d, two into 3s. six into 3p and ten into 3d d) The fourth level has 4 s , 4 p , 4 d and 4f. Ex: Hidrogen. Hidrogen in the first row( 1st level) has only one electron that occupies the 1s orbital. H. 1 s1 * Magnesium Mg ( in the third row ) has 12 electrons. The first two electrons into the 1s orbital, another two into the 2s orbital; six into 2p orbital and the last two into the 3s orbital. Mg. 1s2 2s2. 2p6. 3s2 ** Kripton has 36 e- and in the fourth row, the configuration ; Kr. 1s2 2s,2 2p6. 3s2. 3 p 6 3d10. 4s2 4p6. But since the energy of 3d > 4s, so 2e- filled 4s orbital before 3d. Kr. 1s2. 2s2. 2p6. 3s2. 3p6. 4s2. 3d10 4p6. = ( Ar) 4s2. 3d10. 4p6
Quick Chemistry questions help?
A compound with a molecular mass of 56.104 grams is found to be 86% carbon. The rest of the compound is hydrogen. Find its molecular formula. Select one: a. C2H4 b. C4H8 c. C3H6 d. C6H10 0.86 x 56.104 g = 48.25 g C 48.25 g C / 12.01 g C = 4 56.104 g – 48.25 g C = 7.75 g H = 8 g H Answer is (B) A compound with the empirical formula CH4O2 has a molecular mass of 288.25 grams. What is the molecular formula? Select one: a. CH4O2 b. C2H8O4 c. C4H16Oo d. C6H24O12 CH4O2 = 48.01 288.25 / 48.01 = 6 Multiply 6 through CH4O2 to get C6H24O12 Answer is (D) Which quantity of N2 gas has a volume of 11.2 liters at STP? Select one: a. 1.00 moles b. 2.00 moles c. 14.0 grams d. 28.0 grams PV = nRT P = 1 atm V = 11.2 L n = ? R = 0.08206 L atm/mol K T = 273.15 K n = RT/ PV = 2.00 moles Answer is (B) What is the approximate percent by mass of oxygen in SO3? Select one: a. 20% b. 40% (I chose this one. But I am not too sure.) c. 60% d. 80% S = 32.06 g O3 = 16 x 3 = 48g SO3 = 80.06 O: 48/80.06 x 100% = 60% of O in SO3 Answer is (C)
How do you find Lone Pairs (in Chemistry)?
For the best answers, search on this site https://shorturl.im/ayfQk It actually depends. If an element has been bonded into another element, or simply compound, any pair of electrons that aren't used in chemical bonding is considered a lone pair, the Lewis Dot Structure of the compound will help you determine the number of lone pairs. Example, Water or H2O. Hydrogen has 1 valence electron, while Oxygen has 6. Hydrogen's valence electron will bond to one valence electron of Oxygen, and the second hydrogen will bond to another valence electron of oxygen, leaving oxygen with 4 valence electrons not bonded to any other elements. Since they are 4, we could divide them to 2 since we are talking about pairs, so water has 2 lone pairs.
What is the right way to do organic chemistry for the JEE and the CBSE? Do they need different approaches?
See organic chemistry is that part which i have seen most of the students struggle or are worried about.For me this particularly branch of chemistry came quite easy so i think i am eligible to answer and can share some tips with you prople on how to learn organic.Organic chemistry is that part of chemistry which requires concept clearity and checks your analytical ability in order to solve their problemsTo start with your first chapter in organic will be General Organic Chemistry(GOC) and i think this is by far the most important chapter and in order to have a good command on organic one should master it completely.After this organic would start to come easy start to learn about reaction intermediates,how to propose a mechanism yourself for a particular reaction using conventional mechanisms though it would be difficult in beginning to swallow the reaction mechanisms but with consistent reading and understanding of that mechanism again and again it would definitely come easy.Try to practice a lot of questions particularly on reaction intermediates and their stability,acidic and basic strength of compounds if you are through with this chapter with a proper understanding of concepts then you have a laid a good foundation for your organic next chapter to target would be Hydrocarbons where various reactions along with their mechanisms are proposed try to understand these they would definitely come easy if you have a good command on Goc so it won't be a problem then and try to solve as many questions on the named reactions.If till this chapter you get settled with organic then similar strategy would work for other chapters as well just all you have to do is to have good command on these two chapters and from then you will Start enjoying this beautiful branch of chemistry.And yes there is no need to prepare separately for cbse and jee prepare for jee,cbse would come easy.But yes check out the syllabus for jee organic as there are many topics which are not included in ncert syllabus.Hope it helps.Best of luck for your preparations.