Answer:
yes the Bohr model predicts their spectra accurately
Explanation:
The Bohr model based on one electron system
The atom H has a single atom and the ion Be+3 has a single atom as well and this proves that the Bohr model predicts their spectra accurately .
also apply the equation model to further explain
E = 
z = atomic number
for H the value of z = 1
for Be+3 the value of z = 4
when this values are substituted into the equation above
E =
, E = 
this results show that the energy level of Be+3 is higher than the energy level of H by a factor of 16 , and this shows that their line patterns are similar
Answer:
The average kinetic energy of the molecules in the system
Explanation:
According to the kinetic theory of matter, the particles of a system are in constant motion and collide steadily with one another as well as with the walls of the container.
The average kinetic energy of these molecules in motion is referred to as the temperature of the body. Hence, temperature is a measure of the average kinetic energy of the molecules of a body.
Answer:
5.9405*10^14 Hz
Explanation:
Remember the equation c=v*λ
v=frequency
λ=wavelength
By manipulating the equation, we can get it in terms of frequency:
c/λ=v
Now convert wavelength to meters (1nm=m*10^-9)
5.05*10^-7m (2 decimal places were shifted to make the number between 1-10, hence the -7 exponent)
Use the manipulated equation:
(3.00*10^8)/(5.05*10^-7), which yields:
5.9405*10^14 Hz (or s^-1)
Answer:
From 4.0 moles of hydrogen 2.67 moles of ammonia are produced.
Explanation:
Given data:
Number of moles of hydrogen react = 4.0 mol
Number of moles of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Now we will compare the moles of hydrogen and ammonia from balance chemical equation.
H₂ : NH₃
3 : 2
4 : 2/3×4 = 2.67 mol
From 4.0 moles of hydrogen 2.67 moles of ammonia are produced.
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5