-III I
NH₄⁺
V -II
N₂O₅
I V -II
NaNO₃
-----------------------
-3 +5 +5
Willow is right because the model of atom is not weak. The model of the atom has not changed. It’s just due to new discoveries due to which atom’s model is changing. The particles were always present in the atom but we were not able to discover it.
Answer:
4.48 grams of potassium hydroxide that the chemist must be weighing out.
Explanation:
The pH of the KOH solution = 13
pH + pOH = 14
pOH = 14 - pH = 14 - 13 = 1
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1=-\log[OH^-]](https://tex.z-dn.net/?f=1%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.1 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.1%20M)

1 mole of hydroxide ions are obtained from 1 mole of KOH. Then 0.1 mole of hydroxide ions will be obtained from :
of KOH
![[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}](https://tex.z-dn.net/?f=%5BMolarity%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%28L%29%7D%7D)
Volume of KOH solution = 800 mL = 0.800 L ( 1 mL = 0.001 L)

Moles of KOH = 0.1 M × 0.800 L = 0.08 mol
Mass of 0.08 moles of KOH :
0.08 mol × 56 g/mol = 4.48 g
4.48 grams of potassium hydroxide that the chemist must be weighing out.
Answer:
A
Explanation:
In a galvanic cell, energy is produced by spontaneous chemical processes.
The cathode and anode of this cell will depend on the relative position of the two metals in the electrochemical series.
Aluminium is higher in the electrochemical series so aluminium will be the anode. Silver is lower in the electrochemical series so silver will be the cathode.
Recall that oxidation (electron loss) occurs at the anode while reduction (electron gain) occurs at the cathode.
https://sciencing.com/make-3d-model-atom-5887341.html