Vol.250 before its to much pressure
Answer:
0.48 V
Explanation:
Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)
Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)
Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation
Zinc is the anode while cobalt is the cathode.
E°cell= E°cathode - E°anode
E°cell= -0.28-(-0.76)= 0.48 V
The answer is c. hg (mercury)
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
