Answer: 2.00 V
Explanation:
The balanced redox reaction is:
Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0_{[Al^{3+}/Al]}=-1.66V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAl%5E%7B3%2B%7D%2FAl%5D%7D%3D-1.66V)
![E^0_{[Cu^{2+}/Cu]}=0.340V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D0.340V)
![E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D-%20E%5E0_%7B%5BAl%5E%7B3%2B%7D%2FAl%5D%7D)
Thus the standard cell potential is 2.00 V
Calcium oxide + water = calcium hydroxide.
Magnesium + hydrochloric acid = magnesium chloride + hydrogen.
Sodium hydrogen carbonate + citric acid = Na3C6H5O7(aq)+3H2O(l)+3CO2(g)
Iron + salt water = 4Fe(OH)3 (rust)
Answer:
<h3>The answer is 6 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 15 g
volume = 2.5 mL
We have
We have the final answer as
<h3>6 g/mL</h3>
Hope this helps you
Answer:
After 6sec 5 atoms will be left.
Explanation:
Given data:
Total atoms of Highlinium = 40
Half life = 2 sec
Atoms left after 6sec = ?
Solution:
First of all we will calculate the number of half lives passes during 6sec.
Number of half lives = Time elapsed/ half life
Number of half lives = 6 sec / 2sec
Number of half lives = 3
Now we will calculate the number of atoms left
At time zero = 40 atoms
At 1st half life = 40 atoms/2 = 20 atoms
At 2nd half life = 20 atoms/2 = 10 atoms
At 3rd half life = 10 atoms / 2 = 5 atoms
Thus, after 6sec 5 atoms will be left.
