Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
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Answer:
1000N is needed to be applied.
Explanation:
Machines make doing work easier. They allow us use small effort to carry out work on huge amount of load.
The mechanical advantage of a machine;
(M.A) =load/effort
M.A = 0.6
Load =600N
effort =?
0.6 = 600/effort
effort = 600/0.6
effort = 1000N
Answer:
a. Dipole-dipole bonding
Explanation:
SO2 has dipole-dipole bonding. This is because of the difference in the electronegativities of Sulphur and oxygen. Moreover, the lone pair of electrons on S gives it bent shape with a net dipole unlike CO2 which has a linear shape.( This why CO2 does not have any dipole moment).
So, the correct answer is a.
<span>C is the correct answer. Elements in the periodic table are grouped based on having similar properties. For example, the noble gases are all non-reactive and non-metallic. The electronic structure of an atom is the way the electrons are arranged within it, and this affects where they are located in the periodic table. The number of electrons in an element is the same as its group number in the periodic table (with the exception of Group 0).</span>
Answer:
412.1kJ
Explanation:
For the reaction , from the question -
4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]
The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol
Δ H° Fe = 0
Δ H° O₂ = 0
Putting in the above equation ,
ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0
ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol
- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.
Hence ,
for 1 mol of Fe ,
- 1648.4 kJ/mol / 4 = 412.1kJ
