Answer:
One moves the arm one moves the legs???
Explanation:
Answer:
a. 1.23 V
b. No maximum
Explanation:
Required:
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
If E°cell must be at least 1.10 V (E°cell > 1.10 V),
E°red, cat - E°red, an > 1.10 V
E°red, cat - 0.13V > 1.10 V
E°red, cat > 1.23 V
The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.
Answer:
Find g
Write your answer in simplest radical form
Answer:
41 mL
Explanation:
Given data:
Milliliter of HCl required = ?
Molarity of HCl solution = 4.25 M
Mass of CaCO₃ = 8.75 g
Solution:
Chemical equation:
2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 8.75 g / 100.1 g/mol
Number of moles = 0.087 g /mol
Now we will compare the moles of CaCO₃ with HCl.
CaCO₃ : HCl
1 : 2
0.087 : 2/1×0.087 = 0.174 mol
Volume of HCl:
Molarity = number of moles / volume in L
4.25 M = 0.174 mol / volume in L
Volume in L = 0.174 mol /4.25 M
Volume in L = 0.041 L
Volume in mL:
0.041 L×1000 mL/ 1L
41 mL
