The answer is Ra
Atomic number for Be is 4
Atomic number for Mg 12
Atomic number for Ra 88
Atomic number for Ba 56
It can because Iron is more reactive than Hydrogen and a more reactive metal always replaces a less reactive one in a single replacement reaction
Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
