What is the coordination complex of copper with ammonia as a ligand?
355760
It's [Cu(NH3)6]2+ or hexaamine copper (II)
984431
The oxidation state of Vanadium in the NH4VO3 complex is +5 because oxygen has a -2 charge and NH4 has a +1 charge. 3 oxygens give you a -6 charge from oxygen. -6+1=-5. Since the complex is a neutral complex, vanadium would have to be +5 to match.
1026944
CoCl3.6NH3 yellow 3 [Co(NH3)6]Cl3 \nCoCl3.5NH3 purple 2 [Co(NH3)5Cl]Cl2 \nCoCl3.4NH3 green 1 [Co(NH3)4Cl2]Cl \nCoCl3.4NH3 violet 1 [Co(NH3)4Cl2]Cl \nThe green and violet compounds must have some further difference. \nAll have 6 ligands (coordination number = 6)
113751
Ammonia is a gas at room temp, dissolved in water it it very basic and would have a high pH of around 10 depending on the concentration
913568
Zn(NO2)2; Co(HSO3)3; HgF; Al2(CO3)3; Cs2C2O4; Ba(ClO)2; SnBr2; Mg(HSO3)2.
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
79044
Here is a centralized page of information, with several links along the left side for more information:\nhttp://www.webelements.com/webelements/elements/text/Cu/key.html
803112
the amino group of NH2-R, attacks the carbon of -N=C=S group using the lone pair on nitrogen. The amino group acts as nucleophile. \n        -                              +\n-N-C(NH2-R)=S\nThen the N of isothiocyanate will be protonated to give\n(both - and + are on N's)\n-NH-C(NH-R)=S\n\nHere the H of NH2 group is migrated on to the N of the isothiocyanate group\n\nu can go through the following link\n\nAll the best
671681
You are looking for element #29 - Copper (Cu).
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Candidates include ammonium carbonate, (NH4)2CO3, or polymers where two CO groups get bridged by one NH.
38819
Transition metal.\n\nName: Copper \nSymbol: Cu \nAtomic Number: 29 \nAtomic Mass: 63.546 amu \nMelting Point: 1083.0 °C (1356.15 K, 1981.4 °F) \nBoiling Point: 2567.0 °C (2840.15 K, 4652.6 °F) \nNumber of Protons/Electrons: 29 \nNumber of Neutrons: 35 \nClassification: Transition Metal \nCrystal Structure: Cubic \nDensity @ 293 K: 8.96 g/cm3 \nColor: red/orange
497226
2NH4V03 + H2S04 ---> (NH4)2S04 + 2V02 + H20
328649
I'd imagine it's because the Nitrogen in Ammonia is (or was historically) regarded as the cationic participant with 3 Hydrogen anions ie N3+(H-)3.\n\nWhereas for water the Oxygen is most definitely the Anion O2-\nwith 2 Hydrogen Cations (H+)2.\n\nThe convention for displaying salts generally takes the form Cation Anion.\n\nThis is moot, however, as both compounds are strongly covalent.
457480
(NH4)2CO3 is Ammonium Carbonate
867434
Construct a Lewis structure for NH3 ( Use N as the central atom ):  first: Determine the total number of valence electrons in the molecule.  In this case there are 8.  Represent shared pairs of electrons between the bonded atoms.  There will be one line connecting each atom, with each line representing two electrons, accounding for six of the eight available electrons.  With the remaining two valence electrons, position them on the fourth unbonded side, satisfying the octet rule for Nitrogen.\n\nThe above described Lewis structure gives you the molecular structureTrigonal Pyramidal ( three bonding pairs and one non bonding pair)\n\nRules for predicting molecular shape by VSPER theory:\n\n1.  Determine Lewis structure\n2.  Determine the number of electron pairs (bonded and non bonded) surrounding a central atom.\n3.  Use VSEPR theory to decide what symmetrical three dimensional shape will accommodate all bonded and non bonded electron pairs.\n4.  Classify the shape of the molecule by considering the arrangement of the joined atoms only, ignoring the nonbonded electron pairs.  That is why, although the NH3 molecule  has four pairs of electrons around the central atom, ammonia's shape is trigonal pyramidal.
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
187343
Cu(HCO3)2\n\nThe bicarbonate (hydrogen carbonate) anion has a charge of -1, so there must be two of them to counteract the +2 charge on the Cu.