Answer:
3. 3.45×10¯¹⁸ J.
4. 1.25×10¹⁵ Hz.
Explanation:
3. Determination of the energy of the photon.
Frequency (v) = 5.2×10¹⁵ Hz
Planck's constant (h) = 6.626×10¯³⁴ Js
Energy (E) =?
The energy of the photon can be obtained by using the following formula:
E = hv
E = 6.626×10¯³⁴ × 5.2×10¹⁵
E = 3.45×10¯¹⁸ J
Thus, the energy of the photon is 3.45×10¯¹⁸ J
4. Determination of the frequency of the radiation.
Wavelength (λ) = 2.4×10¯⁵ cm
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm
2.4×10¯⁵ cm = 2.4×10¯⁷ m
Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m
Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:
Wavelength (λ) = 2.4×10¯⁷ m
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
v = c / λ
v = 3×10⁸ / 2.4×10¯⁷
v = 1.25×10¹⁵ Hz
Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.
Answer:
I prob can bc I'm a bad b*tch
lol
Hey there!
We Know that:
2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)
The equilibrium expression for the reaction is:
Kc = [ Zn⁺² ] / [Ag⁺ ]²
Hope that helps!
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
