Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1
POH of a 0.0072 M=-lg(0.0072) = 2.1426675
Answer: The IUPAC name of
is 5-chloro-2-pentyne
Explanation:
1. First select the longest possible carbon chain. For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
2. The longest possible carbon chain should contain all the bonds and functional groups.
3. The numbering is done in such a way that the carbon containing the functional group or substituent gets the lowest number. Triple bond is given priority over substituent halogen.
4. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne.
Thus the IUPAC name of
is 5-chloro-2-pentyne