One
Let's start by stating what we know is wrong. Equilibrium is achieved when the reactants and products have a stable concentration. That makes D incorrect. Equilibrium is not established until about the 6th or 7th second.
The fact that you get any products at all means that the reactants will become products. Just who is favored has to be looked at very carefully. The products start very near 0. They go up until their concentration at equilibrium. When the reach equilibrium, the products have increased to 17. The reactants have dropped from 40 to 27. By a narrow margin, I would say the products are favored.
C is incorrect. There are still reactants left.
E is incorrect. the reactants started out with a concentration of 40. The reaction is not instantaneous. The concentration was highest at 40 or right at the beginning. This assumes that the reactants were mixed and the products were produced and the water/liquid amount has not changed.
B is incorrect. The concentration of the reactants is higher at equilibrium.
A is wrong. It is product favored.
I'm getting none of the above.
Problem Two
AgBr is insoluble (very). You'd have to work very hard to get them to separate into their elemental form. Just putting AgBr in water isn't enough. Lots of heat and lots of electricity are needed to get the elemental form.
I suppose you should pick B. Mass must be preserved. But if you balanced the equation, it would work with heat and electricity.
The answer is hydroxides.
The elements of the group IA are termed as alkali metals, because their hydroxides are alkaline.
Answer:
0.93 mol
Explanation:
Given data:
Number of moles of Na atom = ?
Number of atoms = 5.60× 10²³
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
5.60× 10²³ atoms × 1 mol / 6.022 × 10²³ atoms
0.93 mol
Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :
act as an oxidizing agent.
The oxidation number of N in is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in 
is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means itself get reduced to and it can act as an oxidizing agent.
act as a reducing agent.
The oxidation number of O in is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in 
is Zero (o).
Now, we conclude that the oxidation number increases this means itself get oxidized to and it can act as reducing agent.
