Aluminum, and magnesium are metals. For metals, reactivity decreases as you go from left to right across the periodic table. Atomic number of Al is 13 and of Mg is 12. Hence the least reactive of these two is therefore aluminum.
Magnesium is "HIGHLY FLAMMABLE" carefully take a small piece and hit it with a torch. If its Magnesium it will "Caution, very, quickly burn.
Aluminum will not react to simple flame, it will only melt with enough direct heat.
Magnesium
==========
Atomic Number: 12
Atomic Symbol: Mg
Atomic Weight: 24.305
Electron Configuration: 2-8-2
Aluminum
========
Atomic Number: 13
Atomic Symbol: Al
Atomic Weight: 26.9815
Electron Configuration: 2-8-3
Hope this helps some. Any questions please feel free to ask. Thank you
<span>Quarks are thought to be the basic component of protons and newtons.</span>
The resultant vector is 11√2 km due north east.
<h3><u>Explanation:</u></h3>
The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.
We need to calculate the magnitude and direction separately.
Here firstly for the magnitude,
The magnitudes are both 11 km and they are at right angles to each other.
So, the resultant magnitude = √(11² +11²) km
=11√2 km
Now for the direction, one vector is due north and the other is due east.
So the resultant vector is due north east.
So the final vector is 11√2 km due North-East.
Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA
Answer: 
Explanation:
Given
Charge discharged 
time taken 
Current is given as rate of change of discharge i.e.

Therefore, the average current is 