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rainingspiders

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  1. I know how to find out the molecular geometry and its corresponding bond angles when all the bonds are single bonds, but how much does a double and triple bond affect the bond angles? What if there is more than one double or triple bond? Or if a molecule as a double AND a triple bond? Thanks!
  2. I have been asked to find the +1 state of Ni and Ti so far, but none for the f-orbital groups yet
  3. I'm confused about how to write the electron config for transition metal ions. From what I understand, I would strip away all the highest s-orbital electrons. If the resulting charge is more than the stated ionic charge, then I add the missing electrons onto the d or f orbitals... is this correct? So Ni+ would be [Ar]3d^9? What if the element has an f orbital too, such as Ce? How would I write the charge for the ion for Ce+? Thanks!
  4. wait, 2.1786 x 10^-18 J? I thought the numerator should be 2.1799 x 10^-18 J?
  5. I keep getting n = 3 for the origin, but the answer is n = 6... what am I doing wrong? Question: Given that the wave number of a line in the Paschen series of hydrogen is 9140 cm^-1, what is the origin of the electron (what energy state?) My method: wave number = 1/(wavelength) = 9140 cm^-1 = 91.4 m^-1 lowest state for Paschen series is n = 3, so n-final = 3 Since this is emission... E = (2.1799 x 10^-18)(1/(n-final)^2 - 1/(n-initial)^2) (plank's constant) * (speed of light) * (1/(wavelength)) = (2.1799 x 10^-18)(1/(3)^2 - 1/(initial state)^2) Solving for n-initial... n-initial =3. I know this is wrong, but I don't know what I did wrong. Help?
  6. @organism: what does "relativistic" mean? I'm not sure... the equation my textbook gave me says if the speed is not close to the speed of light, the momentum is mv? So I'm guessing it's a typo then?
  7. I thought I was doing this right until I saw the answer in the back of the book.... is it a typo, or am I doing something wrong? Question: Assuming that the smallest measurable wavelength in an experiment is 0.10 femtometers, what is the maximum mass of an object traveling at 100 m/s for which the de Broglie wavelength is observable? My method: equation: wavelength = (plank's constant) / (momentum) wavelength = (plank's constant) / (mv) 0.10 * 10^-15 m = (6.626 x 10^-34) / (m * (100)) Solving from m... I got 6.6 * 10^-20 kg? But in the back of the book, it says 6.6 * 10^-29 kg? Did I do something wrong? Or did someone typed 9 instead of 0?
  8. Hey, sorry for the late reply. I had midterms this week so I wasn't able to go on the internet that often. Anyways, it turned out I didn't take the spectator ions into account.
  9. Ok, I'll try my best. This is my first time trying to explain Chem so sorry if it's a bit confusing. First, I assigned a different variable for each compound of the equation. The numbers of these variables will be the coefficients of their corresponding compounds. So... HCl = a As2O3 = b NaNO3 = c H2O = d NO = e H3AsO4 = f NaCl = g I then found out how much of each element is on the reactant side and how much of each element is on the product side, with respect to the variables. Using the following equation... # element X on reactant side = # element X on product side For H, we have "a" amount of H and "2d" amount of H, so we get a + 2d amount of h on the left side. We have 2d because for every compound d, we have 2 H's. Do the same on the product side, and repeat for all the elements you have. Doing so gets us... H: a + 2d = 3f Cl: a = g As: 2b = f O: 3b + 3c + d = e + 4f Na: c = g N: c = e I set a variable equal to 1. I chose the variable "a", just because. Doing so gives us a = 1 g = 1 c = 1 e = 1 so... H: (1) + 2d = 3f Cl: (1) = (1) As: 2b = f O: 3b + 3(1) + d = (1) + 4f Na: (1) = (1) N: (1) = (1) The only functions we have that have variables in them are... Equation 1: 1 + 2d = 3f Equation 2: 2b = f Equation 3: 3b + 3 + d = 1 + 4f , which simplifies to 3b + d = 4f - 2 We want to solve the system of equations. We know what f is, so substitute 2b for f into equation 1 and 2. Equation1: 1 + 2d = 3(2b) 1 + 2d = 6b Equation 3: 3b + d = 4(2b) -2 3b + d = 8b - 2 d + 2 = 5b Solve the system of equations for d. 1 + 2d = 6b >>> -5 - 10d = -30b d + 2 = 5b >>>> 12 + 6d = 30 b combines to 7 + 4d = 0 d = 4 plug in d into equation 1, solve for b, then use be to solve for f. Doing so gives us d = 7/4 b = 3/4 f = 3/2 So listing all our coefficients... a = 1 b = 3/4 c = 1 d = 7/4 e = 1 f = 3/2 g = 1 Multiply these values by the least common denominator, which is 4. This gives us... a = 4 b = 3 c = 4 d = 7 e = 4 f = 6 g = 4 Write out the chemical equation with the corresponding coefficients. [latex] 4HCl + 3As2O3 + 4NaNO3 + 7H2O -> 4NO + 6H3AsO4 + 4NaCl [/latex] If we check, we see that the equation balances.
  10. @ Empress: Yeah, I have tried it and it didn't work out for me. However, I just finished solving it, using the algebraic method. Thanks for the help though! Should I post how I solve it here so other people can see?
  11. I'm supposed to balance the following equation: [latex] HCl + As2O3 + NaNO3 + H2O -> NO + H3AsO4 + NaCl [/latex] Anyways, I got stuck stuck here... [latex] HCl + As2O3 + NaNO3 + 3H2O -> NO + 2H3AsO4 + NaCl [/latex] No matter what I do I couldn't get H and O to balance out at the same time! When I balance O, the H is out of balance, and when I balance H, O is out of balance! Augh! How do I do this? Ok, I have solved it to here... [latex] 2HCl + As2O3 + 2NaNO3 + 2H2O -> 2NO + 2H3AsO4 + 2NaCl [/latex] There are 11 O on the right and 10 O on the left. ... Now what?
  12. I'm supposed to balance the following equation: [latex] HCl + As2O3 + NaNO3 + H2O -> NO + H3AsO4 + NaCl [/latex] Anyways, I got stuck stuck here... [latex] HCl + As2O3 + NaNO3 + 3H2O -> NO + 2H3AsO4 + NaCl [/latex] No matter what I do I couldn't get H and O to balance out at the same time! When I balance O, the H is out of balance, and when I balance H, O is out of balance! Augh! How do I do this? I realized that I might have posted this on the wrong topic of the forum. Is it possible for me or a moderator to delete this topic? I am going to post it again on the homework help section. Sorry!
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