If an atom contains 13 protons, then it has <u>13 electrons.</u>
Answer:
The magnifying power of this telescope is (-60).
Explanation:
Given that,
The focal length of the objective lens of an astronomical telescope, 
The focal length of the eyepiece lens of an astronomical telescope, 
To find,
The magnifying power of this telescope.
Solution,
The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :


m = -60
So, the magnifying power of this telescope is 60. Therefore, this is the required solution.
Answer:
7.60× 10^6 V/m
Explanation:
electric field strength can be determined as ratio of potential drop and distance, I.e
E=V/d
Where E= electric field
V= potential drop= 74.0 mV= 0.07 V
d= distance= 9.20 nm = 9.2×10^-9 m
Substitute the values
E= 0.07/ 9.2×10^-9
= 7.60× 10^6 V/m
Isotope has 57.40% abundance
the other one 100% - 57.40% = 42.60%
mass of 121 (% abundance) + mass of 123 (% abundance) = total mass (atomic weight)
(120.9038 amu)(0.5740) + (x amu)(0.426) = 121.76 amu
x = 122.9136
hope this help