Ans: Moles of Fe(OH)2 produced is 5.35 moles.
Given reaction:
Fe(s) + 2NiO(OH) (s) + 2H2O(l) → Fe(OH)2(s) + 2Ni(OH)2(aq)
Based on the reaction stoichiometry:
1 mole of Fe reacts with 2 moles of NiO(OH) to produce 1 mole of Fe(OH)2
It is given that there are:
5.35 moles of Fe
7.65 moles of NiO(OH)
Here the limiting reagent is Fe
Therefore, number of moles of Fe(OH)2 produced is 5.35 moles.
Answer:
H⁺(aq) + OH⁻(aq) ⇒ H₂O(l)
Explanation:
Let's consider the molecular equation that occurs when aqueous solutions of perchloric acid and potassium hydroxide are combined. This is a neutralization reaction.
HClO₄(aq) + KOH(aq) ⇒ KClO₄(aq) + H₂O(l)
The complete ionic equation includes all the ions and molecular species.
H⁺(aq) + ClO₄⁻(aq) + K⁺(aq) + OH⁻(aq) ⇒ K⁺(aq) + ClO₄⁻(aq) + H₂O(l)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
H⁺(aq) + OH⁻(aq) ⇒ H₂O(l)
A) a good hypothesis to be put in question for the experiment would be “Fluorescent light can fade color photographs quicker than direct sunlight can.”
B) the controlled variables, or things that will need to be kept constant throughout the experiment, would be the amount of light on each picture, the amount of ink on each picture, the type of ink on each picture, the size of each picture, and the time of day of the light for each picture.
They have 6 electron in their outer orbital. That means they want 2 electrons to be stable. Since metals have the tendency to give electron oxygen receives 2 electrons. The answer is -2
Answer:
0.43.
Explanation:
In order to solve this question we will need the an equation called Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equation can be represented by the reaction below;
pH= pKa + log ( [ A^- ] / [ HA] ).
Where HA is the acetic acid and A^- is the Acetate ion
We are given the pH value to be = 4.38 and the ka to be = 1.76×10^–5. So, we will use the value for the ka to find the pKa through the formula below.
pKa = - log ka.
Therefore, pKa = - log( 1.76×10^–5).
pKa= 4.75 + log
So,
4.38 = 4.75 + log ([ A^-] / [HA]).
4.38 - 4.75 = log ( [ A^- ] / [ HA] ).
( [ A^- ] / [ HA] ) = 10^- 0.37.
( [ A^- ] / [ HA] ) = 0.42657951880159265.
( [ A^- ] / [ HA] )= 0.43.
