A cathode is the location in an electrolytic cell where reduction reactions take place.
An anode is the location in an electrolytic cell where oxidation reactions occur.
An electrolyte solution is any substance containing free ions that make the substance electrically conductive.
<span>An external electrical energy source like a battery or a transformer is used to drive the non-spontaneous reaction.</span>
The correct answer is 0.06857 moles.
C₆H₁₂O₆, that is, glucose has six carbons, twelve hydrogens, and six oxygen atoms. The atomic weight of C, H and O are as follows:
Six atoms of carbon = 6 × 12.01 g = 72.06 g
Twelve atoms of hydrogen = 12 × 1.008 g = 12.096 g
Six atoms of oxygen = 6 × 16.00 g = 96.00 g
So, the molar mass of C₆H₁₂O₆ is 72.06 g + 12.096 g + 96.0 g = 180.156 g.
It can also be written in the form as 180.16 g of C₆H₁₂O₆ is equal to 1 mole of C₆H₁₂O₆or 180.16 g/mole (as the molar mass)
Now, there is a need to find moles of 12.354 grams of C₆H₁₂O₆. So, the final conversion is:
12.354 g C₆H₁₂O₆ × 1 mole of C₆H₁₂O₆ / 180.16 g C₆H₁₂O₆
= 0.06857 moles
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
Answer is in a pho
to. I can only upload it to a file hosting service. link below!
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Answer:
Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.
Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g
Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol
Explanation: To calculate concentration of quinine, Cb is as follow
Va*Ca=Vb*Cb
∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol
