36.5 grams per mole. According to the periodic table, the atomic mass of hydrogen is 1 and the atomic mass of chlorine is 35.5 grams per mole. The molar mass of HCl would then be calculated.
Male rats were dosed with (14)C-ethylenediamine (EDA) at 5, 50, or 500 mg/kg by oral, endotracheal, and iv routes. Urinary excretion was the primary route of elimination accounting for 42 to 65% of the admin radioactivity. Calculate the reduced mass of HCl molecule given that the mass of H atom is 1.0078 amu and the mass of Cl atom is 34.9688 amu. Note that 1 amu = 1.660565.10 -27 kg. The mass of one hydrogen atom is 1gm and the atomic mass of chlorine is 35.5 gm. The atomic mass of HCl will be 35.5+1 =36.5 gm HCl is hydrochloric acid and it has a boiling point of -85.05.
Solutions to select questions can be found online.
5.7
Calculate the reduced mass of HCl molecule given that the mass of H atom is 1.0078 amu and the mass of Cl atom is 34.9688 amu. Note that 1 amu = 1.660565*10-27 kg.
Then (hat{L}^{2}=0) and (hat{L_z}=0). Given that the values for (l) and (m) are the same as above, the answers would also be the same.
The reason why both answers are the same is that the operators for angular momentum only act on the angular part of the wave function. Since only the quantum number (n) varied between these two states, the angular momentum eigenvalues did not change.
5.38
Apply the angular momentum operator in the x direction to the following functions ((Y(theta,phi))).
(dfrac{5pi}{4} + 7exp(pi^2) )
(3pi sin(theta) )
(dfrac{3}{2}cos(theta)exp(iphi))
Solution
Let us begin by stating the angular momentum operator in terms of (theta) and (phi).
Determine the unnormalized wave function (psi_circ big(xbig)) given that (hat{a}_- = 2^{-1/2}big(hat{x}+ihat{p}big) ) and that (hat{a_-}psi_circ = 0) Then find the unnormalized wave function for (psi_1big(xbig)) using (hat{a}_+).
Solution
It was given that (hat{a}_-psi_circ = 0), so substituting in (hat{a}_-) so we know
Find the reduced mass of an electron in a Tritium atom. Set the mass of the Tritium to be (5.008267 times 10^{-27}, kg). Then find the value of the Rydberg constant for the Tritium atom.
Solution
To solve, use the reduced mass equation, and for mass 1 enter the mass of the electron, and for mass 2 enter the mass of the Tritium atom:
[mu = dfrac{m_1m_2}{m_1+m_2} nonumber ]
For which one attains a value of (mu = 9.1077 x 10^{-31} kg)
5.46
The mass of a deuterium atom is (3.343586 times 10^{-27}; kg). First calculate the reduce mass of the deuterium atom. Then using the reduced mass calculated find the Rydberg constant for a deuterium atom.
Solution
(mu) = reduced mass
[mu_{deuterium} = dfrac{(9.109390 times 10^{-31} kg) ( 3.343586times 10^{-27}kg)}{(9.109390 times 10^{-31} kg + 3.343586times 10^{-27}kg )} nonumber ]
[mu_{deuterium} = 9.106909 times 10^{-31}kg = 0.9997277 m_e nonumber ]
What is the ratio of the frequency of spectral lines of C-14 that has been ionized 5 times and C-12 that has been ionized 5 times?
Solution
Carbon that has been ionized 5 times is a hydrogen like ion, so we can use the Bohr model to find the desired ratio.
[E = dfrac{uZ^2e^4n^2}{8ε_0^2h^3c} nonumber ]
gives the placement of spectral lines. The coefficient of n2 is proportional to the frequency of these lines, so the ratio of EC-14/Ec-12 will give the ratio of frequency of the lines. The only difference between these two isotopes is the reduced mass u. So the problem reduces to uC-14/ uC-12. Mass in amu is used below. me = mass of electron = 5.4858*10-4 amu.
[mu_{C-14} = dfrac{m_em_{c-14}}{m_e + m_{c-14}} = dfrac{(14.003)(5.4858 times 10^{-4})}{14.003 + 5.4858 times 10^{-4}} = 5.485585 times 10^{-4} nonumber ]
[mu_{C-12} = dfrac{m_em_{c-12}}{m_e + m_{c-12}}= dfrac{(12)(5.4858 times 10^{-4})}{12 + 5.4858 times 10^{-4}} = 5.485549 times 10^{-4} nonumber ]
Calculate the Rydberg constant for a deuterium atom and atomic hydrogen given the reduced mass of a deuterium atom is (9.106909 times 10^{-31} kg) and the reduced mass of hydrogen is (9.104431 times 10^{-31} kg). Compare both of these answers with the experimental result ((109677.6 cm^{-1})). Then determine the ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium.
Solution
The Rydberg constant is found using
[R_H=dfrac{me^4}{8epsilon_o^2 ch^3} nonumber ]
For a deuterium atom
[R_H=dfrac{(9.104431 times 10^{-31}kg)(1.602 times 10^{-19} C)^4}{8(8.854 times 10^{-12} dfrac{F}{m})^2(2.998 times 10^{8} dfrac{m}{s})(6.626 times 10^{-34} J cdot s)^3} nonumber ]
[R_H=109707.3 cm^{-1} nonumber ]
This is different by ( 2.7 times 10^{-2}%).
The ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium is equivalent to the ratio of the Rydberg constants we just found.
[R_H=dfrac{(9.106909 times 10^{-31}kg)(1.602 times 10^{-19} C)^4}{8(8.854 times 10^{-12} dfrac{F}{m})^2(2.998 times 10^{8} dfrac{m}{s})(6.626 times 10^{-34} J cdot s)^3} nonumber ]
[R_H=109677.5 cm^{-1} nonumber ]
This is different by ( 9.1times 10^{-5}%).
For a hydrogen atom
5.46
Find the reduced mass of HCl where the mass of hydrogen in 1 amu and the mass of chloride is 35 amu.
Solution
[mu = dfrac{m_1 times m_2}{m_1 + m_2} nonumber ]
[mu = dfrac{(1.00)(35.00)}{36.00} 1.603 times 10^{-27} kg = 1.558 times 10^{-27} kg nonumber ]
Molar mass of HCl = 36.46094 g/mol
Convert grams Hydrochloric Acid to moles or moles Hydrochloric Acid to grams
Molecular weight calculation: 1.00794 + 35.453
Symbol
# of Atoms
Chlorine
Cl
35.453
1
97.236%
Hydrogen
H
1.00794
1
2.764%
Density Of Hcl
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
Molar Mass Of Hcl
The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.