Answer:
a. 750Hz, b. 4.0ppm, c. 600Hz
Explanation:
The Downfield Shift (Hz) is given by the formula
Downfield Shift (Hz) = Chemical Shift (ppm) x Spectrometer Frequency (Hz)
Using the above formula we can solve all three parts easily
a. fspec = 300 MHz, Chem. Shift = 2.5ppm, 1MHz = 10⁶ Hz, 1ppm (parts per million) = 10⁻⁶
Downfield Shift (Hz) = 2.5ppm x 300MHz x (1Hz/10⁶MHz) x (10⁻⁶/1ppm)
Downfield Shift = 750 Hz
The signal is at 750Hz Downfield from TMS
b. Downfield Shift = 1200 Hz, Chemical Shift = ?
Chemical Shift = Downfield shift/Spectrometer Frequency
Chemical Shift = (1200Hz/300MHz) x (1ppm/10⁻⁶) = 4.0 ppm
The signal comes at 4.0 ppm
c. Separation of 2ppm, Downfield Shift = ?
Downfield Shift (Hz) = 2(ppm) x 300 (MHz) x (1Hz/10⁶MHz) x (10⁻⁶/1ppm) = 600 Hz
The two peaks are separated by 600Hz
Answer:
Both have the same amount of particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.
This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.
Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.
This sounds very much like a chicken-egg problem.
The first thing that formed must be hydrogen nuclei. The only other alternative is that the atom was created instantly, and the nuclei sprang forth at the same time as the atom, meaning that neither was technically first. The logic is that an atom can’t form without a nucleus, but it theoretically could be created instantly.
Answer:
60g
Explanation:
Moles = mass / mollar mass
Mass = Moles x mollar mass
Mass = 2.50 x 24
Mass = 60g
Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
