For any number (a + bi), its conjugate is (a - bi), so the real part stays the same, and the imaginary part is negated.
For this one, conjugate of [-3 - 9i] is: -3 + 9i
PO43-
No, the equation is: HPO4-2 → H+ + PO4-3 The HPO4-2 and PO4-3 are conjugate acid base pairs.
H3PO4 (aq) + H2O (l) ---> 2H3O+ (aq) + PO4-3 (aq)donor acid + acceptor base ---> conjugate acid + conjugate basethe answer above is wrongto form a conjugate, the ion H2PO4 - must lose a hydrogen ion H+i.eH2PO4 - -H+ = HPO4 2-(conjugate base)
H3PO4 (aq) + H2O (l) ---> 2H3O+ (aq) + PO4-3 (aq)donor acid + acceptor base ---> conjugate acid + conjugate basethe answer above is wrongto form a conjugate, the ion H2PO4 - must lose a hydrogen ion H+i.eH2PO4 - -H+ = HPO4 2-(conjugate base)
HNO2 conjugate acid = one more hydrogen conjugate base = one less hydrogen
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
For any number (a + bi), its conjugate is (a - bi), so the real part stays the same, and the imaginary part is negated.For this one, conjugate of [-3 - 9i] is: -3 + 9i
PO43-
0 + 5i Its complex conjugate is 0 - 5i
3-2j.
No
phosphate
It is 3 minus 2i
H3PO4 (aq) + H2O (l) ---> 2H3O+ (aq) + PO4-3 (aq)donor acid + acceptor base ---> conjugate acid + conjugate basethe answer above is wrongto form a conjugate, the ion H2PO4 - must lose a hydrogen ion H+i.eH2PO4 - -H+ = HPO4 2-(conjugate base)
No, the equation is: HPO4-2 → H+ + PO4-3 The HPO4-2 and PO4-3 are conjugate acid base pairs.
H3PO4 (aq) + H2O (l) ---> 2H3O+ (aq) + PO4-3 (aq)donor acid + acceptor base ---> conjugate acid + conjugate basethe answer above is wrongto form a conjugate, the ion H2PO4 - must lose a hydrogen ion H+i.eH2PO4 - -H+ = HPO4 2-(conjugate base)
H3PO4 (aq) + H2O (l) ---> 2H3O+ (aq) + PO4-3 (aq)donor acid + acceptor base ---> conjugate acid + conjugate basethe answer above is wrongto form a conjugate, the ion H2PO4 - must lose a hydrogen ion H+i.eH2PO4 - -H+ = HPO4 2-(conjugate base)