Answer is: Rutherford demonstrate that J.J Thompson's Plum Pudding model was not accurate.
Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model, because it shows that most of the atom is empty space.
The balanced equation for the neutralisation reaction is as follows
2HNO₃ + Ba(OH)₂ ---> Ba(NO₃)₂ + 2H₂O
stoichiometry of HNO₃ to Ba(OH)₂ is 2:1
number of Ba(OH)₂ mol present - 0.108 mol
1 mol of Ba(OH)₂ neutralises 2 mol of Ba(OH)₂
therefore 0.108 mol of Ba(OH)₂ neutralises - 2 x 0.108 mol = 0.216 mol of HNO₃
Answer:
So sorry if I was wrong but I think it's B. Because from the source states,
https://socratic.org/questions/which-group-on-the-periodic-table-is-the-least-reactive-why
"The least reactive elements are those who have a full outermost valence shell ie they have 8 electrons in the outer shell so elements such as helium, neon, radon or the transition elements."
Answer:
1) Increase
2) Decreases
3) increases
4) Increase
Explanation:
These questions can only be answered by considering the principle which states that, "When a constraint such as a change in concentration, pressure or volume is imposed on a reaction system in equilibrium. The system will readjust itself in order to annul the constraint."
Now, if more reactants are added, the equilibrium position will shift towards the right, If more products are added, the equilibrium position will shift to the left.
Similarly, the removal of H2S causes the O2 concentration to increase since the equilibrium position now shifts to the left.
Also, addition of O2 causes H2S to be removed as the equilibrium moves to the right.
Answer:
in the attached image is the reaction mechanism.
Explanation:
The first reaction (reaction 1) shown in the attached image is the Wolff-Kishner reduction, which is characterized when the carbonyl is reduced to an alkane in the presence of a hydrazine and a base. In reaction 1, the aldehyde reacts with hydrazine to produce oxime. This mechanism begins with the attack of the amine on the carbonyl group. Proton exchange happens and the water leaves the molecule.
In reaction 2, the KOH is deprotoned in nitrogen and organized to form the bond between the nitrogen molecule. this deprotonation releases the nitrogen gas
