For a neutralization reaction, the value of q(heat of neutralization) is doubled when the concentration of only the acid is doubled.
A neutralization reaction is a reaction in which an acid reacts with a base to yield salt and water. Ionically, a neutralization reaction goes as follows; H^+(aq) + OH^-(aq) ------> H20(l).
The heat of neutralization (Q) of the system depends on the concentration of the solutions. Since Q is dependent on concentration, if the concentration of any of the reactants is doubled, more heat is evolved hence Q is doubled.
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Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Answer:
All atoms heavier than barium
Explanation:
In the periodic table, elements are divided into blocks. We have the;
s- block elements
p- block elements
d- block elements
f- block elements
However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.
Hence elements heavier than barium all possess f-orbitals.
Bromine (Br) is the only element listed that is a diatomic element. I believe that Bromine is going to be your answer.
Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
