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chemistry

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The particle wavelength of your lecturer in 3.091 on his way from the office to 10-250 (200 yards, 1 yard = 0.9144 m) has been determined to be 4.045 x 10 -36m. Taking his body weight to be 178 lbs (1 lb = 0.4536 kg), how long does it take him to get to his class? (answer in seconds)

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    wavelength = h/mv
    Solve for v. Then
    distance = rate x time
    Substitute and solve for time.

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    pls the formula :)

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    thanks bob it worked!

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    can u help me KS

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    Hi, Menes. I an getting a very strange answer, which is incorrect. I must be using the wrong formula. Can you assist. Thanks.

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    Okay i am going to list down all the answers i got so far, please assist me in those which i left, if you have any:

    1)156.675158119
    2)5
    3)90.1775147929
    4)3s,3p,4s,3d,5s,5p,4f
    10,2,14,6
    5)-777
    2.86e-10
    6)8.88
    dont know what part b is
    7)O2, O2-2
    8)tetrahedral,2,1,1,no,no
    9)LiF don't know the reason
    10)dont know part a
    6.3e4
    11)BN,AlAs, C,GaAs,GaP
    12)
    13)1.04e-3,p-type
    14
    15
    16
    17)Sn < Ge < Si
    18
    19) 6
    20
    21) B
    22)1.8e-6
    23)0.000258
    24)
    25
    26
    27
    28)A is wrong, 3320, thermoplastic no covalent bonds
    29
    30
    31
    32)-0.0453
    14.0453
    33)4.77
    10.76
    1
    34) hydrogen bond

    Pls help me get the rest ones ASAP

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    Hi KS, Thanks. 6b) 3.22*10^5

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    You are welcome, thanks. I forgot to convert from KJ/mol to J/mol!
    If you get anything else, pls do share on this page!

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    This is what I did with No. 12 Gallium Nitride.
    A) I said No.
    B) Wavelength = h*c/E_g = 6.626*10^-34 x 3*10^8/ 3.2 x 1.602 * 10^-19
    = 0.388 *10^-6
    This is a long shot. Maybe someone can pick up on it.

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    28 is A and D

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    thanks for 28 it's correct, as for 12, a is yes and b is 3.88e-7 I got them correct. Thanks again. feel free to brainstorm for other questions.

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    thanks KS, 31 is A

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    9. b) I think the answer is "Both molecules are ionic but the higher melting point has a larger cohesive energy". I ruled out some of the choices, LiF has a strong lattice energy. There are a lot of choices to choose from. This is what I think.

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    thanks kim, thanks hil, both your answers are right. Good going!

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    Hi. KS, I have a few more to go. Only one chance left. The graphs, the one on gold, x-ray tube. If you can fill in what you have. 9b is correct.21. is b and both glasses are weaker. Thanks

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    Thanks man, i am telling you all i know.. don't worry, if you get more than 60 overall you will receive your certificate (like i did in bio despite failing in the final, they don't mention your score in the certificate) I don't know anything about gold or the graphs yet. rest all leads brought me nowhere, will keep you posted if I find something, we have one more day.

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    I am also telling the answers of those i myself got wrong in my previous tries. 10 a is 2.42e-6 I got it wrong

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    try the stereoisomers one q27, looks pretty straightforward.

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    Hi guys , , the answer for 6 b) is 3.22e5 .

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    Yeah hi. Hil told us that

    can someone solve the lead solubility and tell us the end result please? Here are the steps:
    "This is simple Stoichiometry problem

    Write down PbI2 -> Pb2+ + 2I-

    Therefore

    Ksp = [Pb2+][I-]^2

    From the stoich you can derive that

    Ksp = Cs(2Cs)^2 = 4Cs^3

    Hence Cs = (Ksp/4) ^ 1/3

    Then multiply the Molar Weight to get grams / Liter unit that you like

    The second problem is common ion effect

    First see is the added ion concentration more than or less than your Cs.

    0.1 seems to be >>>> than Cs so common ion will happen

    Ksp = Cs'(0.1)^2 = 0.01Cs'

    Hence

    Cs' = Ksp/0.01

    Then just multiply by Molar weight again to convert to g/L units .

    Note use the Molar Weight of PbI2 !!!!"

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    Alright i got it
    24)0.558
    0.000327

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    Can somebody pls help me with 27 , 29 and 30 ? I'd be really grateful ,

    Thankyou

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    The answers for question 20 are

    Structure A ,

    Increase ,

    Decrease .

    Hope it helps . .

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    Hi, KS and Kim. Can you help me with No. 18 electrical measurements. Metal is two orders of magnitude higher at 836 degrees than at 499 degrees C. What is the energy formation for this metal. Thanks.

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    Hi, KS and Kim. Can you help me with No. 18 electrical measurements. Metal is two orders of magnitude higher at 836 degrees than at 499 degrees C. What is the energy formation for this metal. Thanks.
    In(100)= H_v/8.314(1/772.15-1/1109.15)

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    On no. 25. I believe the highest density is Coesite. The others are tricky. Does anyone know. On 26, phases for Cu at 95wt.% I think is 2. At 15wt.% for Cu, I think is 2. For no. 29, I think linear structure has a larger molar volume. Branched structure a higher glass transition temperature. Part c is a toss up. I think it is branched because the greater amount of electrostatic interactions in the linear chain.

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    9a is LiF

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    pls help me with 27, 29 and 30 .

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    ¹14 is 1.42e9,
    2.36e-10

    ¹15 is 7.23e-11,
    2.48e-11

    ¹16 is 6.89


    ¹20 is structure A, decrease, increase
    Zm is wrong

    ¹29 is
    branched structure,
    branched structure,
    Linear, because the branches can become more entangled than the linear chain.

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    Thanks Tania. I was making calculation errors and unsure of what they were looking for on the gold problem. I am having he same problem with no 18. I know the formula but getting the wrong answer. Temp = 836 C and 499 C
    In(100)=H_v/8.314(1/772.15-1/1109.15)
    Thanks.

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    Is the answer in number 29 correct? Linear is less viscous? isn't it suppose to be that branched be less viscous?

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    anyone up for 16?

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    Estimate the fraction of copper-rich phase at 900 C and an overall composition of 80 wt. % Cu. Have anyone solve that?

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    No.29 is correct. No. 18 can you check my math.
    ln(100)=H_v/8.314*(1/772.15-1/1109.15)
    ln(100)= H_v/8.314*0.000393
    H_v= ln(100)*8.314/0.000393
    H_v=ln(100)*21155.2163)
    H_v=97423.37 joules
    H_v=97.42337 kilo joules
    Thanks.

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    ^ seems to be correct make sure that your queestion says double magnitude and those values for temperature are all Kelvin

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    The answer for 18 is 92.8892497896....
    Any help for No.30(K SURFACE ENERGY)?

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    I get that why linear is less viscous than Branched, i forgot that this question is about polymers and not short chain hydrocarbons...

    Has anyone have answers for the Stereoisomer thing. I am very confused with it cause of the wording "number of possible stereoisomers"

    Cause if that was the question then i would just use the formula 2^n where n is the number of chiral centers!!! For this the answer is 2^2 = 4

    But then again if it was really asking number of possible optically different stereoisomers than the answer is only 3 since the other one is only similar to the other due to the plane of symmetry, i bet it was called a meso compound back in the tartaric acid lecture !!!

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    for K Surface Energy just do the following steps

    Potassium is BBC hence 2 atom per lattice

    Use Atomization Enthalphy from periodic table -- Divide this by avogadros -- to get per atom -- then remember that in BBC u have 8 surrounding atoms, -- so divide that value again by 8

    That now will be energy / bond !!!

    Then now calculate the area of the plane (110)

    Just get lattice constant (a) using (2*Molar Volume / Avogadros) ^ 1/3

    Theen solve the area using = a*sqrt(a)

    Then now use your energy in Joules / bond and divide it by the area your calculated.

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    Alright i tried my theory and my understand about the meso compound situation is correct!!!

    So if you see an axis of symmetry remove 1 from the 2^n formula

    :)

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    No. 25 This is my answer: lowest density: cristobalite;
    next lowest: tridymite; next lowest: beta quartz; next lowest: alpha quartz; highest density: coesite.
    Please verify.

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    25) tridymite,crystobalite, beta-quartz, alpha-quartz,coesite.
    cristobalite and liquid have approximately the same density

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    guys how about this question?

    What is the minumum acceleration potential (V) that must be applied to a beam of free electrons for diffraction to occur in a crystal of niobium (Nb)? Express your answer in volts.

    I've answered it by both 2663V and 2.7x10^3 it says im wrong. What is the ryt answer?

    My Solution was Nb is BBC so i find highest plane (110) that gives max wavelength. since h+k+l = even.

    Also lambda max = 2d

    d = a/sqrt(h^2 + k^2 + l^2)
    h = 1, k = 1, l =0 from (110)
    a = (2 * Molar Volume / Avogadros )^2

    Then substitute lamda max to

    Volts = h*c/(e*lambda max) = 2667 Volts

    I tried 2.7 x 10^3 volts i am still wrong :( any suggestions?

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    For above my a
    Sorry Typo
    a = (2 * Molar Volume / Avogadros)^ (1/3)

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    for Nb correct answer is 6.89 Volts.
    Thanks alot ZM for posting the wrong answer. Next time please double check before sharing your results.
    I tried the steps anonymous posted for K surface energy from bananas but its giving me the wrong result. Anyone else can give a try?

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    Im extremelyy sorry KM and others, i just checked. I was a mistake , I had no such intension . Just wannad to help.
    The correct ans. for stereoisomers is

    3 , 4 , 4 . I got it wrong .

    Apologies. . :(
    Thank you .

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    its alright your intentions were good i can see that so your apology is accepted. sorry for being harsh when all you did was trying to help. i met up with someone who got 94 in the finals and he was kind enouigh to give me some answers

    26)1, 3, 0.57; 27 (successively)
    27)3,4,4
    30)9.3e-6

    Thanks to all that helped! couldn't have made it this far without your help. Hope we can assist each other another time :)

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    :) , , agreed with KS , thanks alot everyone for the assistance ! God bless you all . .
    Good Luck :) . .

    Regards.

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    I had 2 wrong attempts of 18 can someone help me with this
    Please
    On the basis of electrical measurements, it is concluded that the equilibrium vacancy concentration in a given metal is two orders of magnitude higher at 849oC than at 492oC. What is the energy of vacancy formation (ΔHv in kJ/mol) for this metal?

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