Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
Answer:
e is the correct one , hope I helped u
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
Answer:
[Ne]3s2
Explanation:
ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.
considerando la configuración electrónica más externa de cada una de las especies mostradas;
para la primera configuración, ns2 np6 corresponde a un gas noble.
para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.
para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.
para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6
what is the answer to what there is nothing here
