Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Molar mass of CuSO4 * 5 H2O
= 63.546 + 32 + 16*4 + 5*18
= 249.546 g/mol
Mass of water in that formula: 5 * 18 = 90 g/mol
Percent by mass of water = 90 / 249.546 = 36%
<span>So, 36% of your 8.22 g is water. 0.36 * 8.22= 2.95 g of water
</span>
The solubility equilibrium of 
:
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
Answer:
The hydroxyl group
Explanation:
The molecular formulas of the above alcohols are
CH₃CH₂-OH
CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂-OH
CH₃-OH
CH₃CH(CH₃)CH₂CH₂-OH
The functional group that is characteristic of all alcohols is the hydroxyl group (-OH).
