Answer:
the stoichiometric coefficient for cobalt is 3
Explanation:
the unbalanced reaction would be
Co(NO₃)₂+ Al → Al(NO₃)₃ + Co
One way to solve is to build a system of linear equations for each element (or group as NO₃) , knowing that the number of atoms of each element is conserved.
For smaller reactions a quick way to solve it can be:
- First the Co as product and as reactant needs to have the same stoichiometric coefficient
- Then the Al as product and as reactant needs to have the same stoichiometric coefficient
- After that we look at the nitrates . There are 2 as reactants and 3 as products . Since the common multiple is 6 then multiply the reactant by 3 and the product by 2.
Finally the balanced equation will be
3 Co(NO₃)₂+ 2 Al → 2 Al(NO₃)₃ + 3 Co
then the stoichiometric coefficient for cobalt is 3
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
Answer:
molecule. The smallest part of a compound is the molecule. A molecule retains all the properties of that compound.
Explanation:
thanks me later
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
