Answer:
λ = 0.5×10⁻⁵ m
E = 39.78×10⁻²¹ J
The given radiation is gamma ray.
Explanation:
Given data:
Frequency of radiation = 6.0 ×10¹³ Hz
Wavelength of radiation = ?
Type of radiation = ?
Energy of radiation = ?
Solution:
Formula:
speed of light = wavelength × frequency
3 ×10⁸ m/s = λ × 6.0 ×10¹³ Hz
Hz = s⁻¹
λ = 3 ×10⁸ m/s / 6.0 ×10¹³ s⁻¹
λ = 0.5×10⁻⁵ m
Energy of radiation:
E = hf
h = planck's constant
f = frequency
E = 6.63×10⁻³⁴ j.s × 6.0 ×10¹³ s⁻¹
E = 39.78×10⁻²¹ J
The given radiation is gamma ray.
Answer:
Cyclobutane is a cycloalkane and organic compound with the formula (CH2)4.
Explanation:
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Answer: yo sorry this a hard one
Explanation:
bro
The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.
