The number of <span>neutrons that are present in its nucleus is 18. The correct option among all the options that are given in the question is the second option or option "B". Mass number minus the atomic number gives the number of neutrons. I hope that this is the answer that has come to your help.</span>
Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:
Voltage in this case is the energy dissipated so
Using the equation to find capacitance
F
C= 5.32 uF because u is the symbol for micro that is equal to 
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Answer: • using beaker tongs to handle the hot beaker.
• checking the beaker for chips prior to heating on the hot plate.
• Turning off the hot plate after use
Explanation:
The options that will ensure laboratory safety during the experiment will be:
• using beaker tongs to handle the hot beaker.
• checking the beaker for chips prior to heating on the hot plate.
• Turning off the hot plate after use.
We should note that the beaker tongs are simply used in the holding of the beakers that have hot liquids in them. Also, it s vital for the hot plate to be turned off after its use so as to prevent accident.
Answer:
a. metallic bond
b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
c. due to the presence of free electrons in its outer energy levels
