Answer:
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
Answer:
0.135 mole of H2.
Explanation:
We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:
Mass of Mg = 3.24 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 3.24/24
Mole of Mg = 0.135 mole
Next, we shall write the balanced equation for the reaction. This is illustrated below:
Mg + 2HCl —> MgCl2 + H2
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.
Thus, 0.135 mole of H2 can be obtained from the reaction.
Dmitri Mendeleev
<span> Dmitri Mendeleev and Lothar Meyer individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass,
certain sets of properties recur periodically.</span>
Answer:
where is the answer options because it sounds like I need some
Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
