Yes mixing salt with pepper change
Answer:
An orbital is a region in space where there is a high probability of finding an electron.
Explanation:
The orbital is a concept that developed in quantum mechanics. Recall that Neils Bohr postulated that the electron occupied stationary states which he called energy levels. Electrons emit radiation when the move from a higher to a lower energy level. Similarly, energy is absorbed by an electron to move from a lower to a higher orbit.
This idea was upturned by the Heisenberg uncertainty principle. This principle state that the momentum and position of a particle can not be simultaneously measured with precision.
Instead of defining a 'fixed position' for the electron, we define a region in space where there is a possibility of finding an electron with a certain amount of energy. This orbital is identified by a set of quantum numbers.
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
Answer:
0.2042 M is the original concentration of 
(aq) in the titrating solution.
Explanation:
Mass of 
= 0.217 g
Moles of 
1 mole of have 2 mole of As and 1 mole of 
have 1 mole of As.
So, from 1 mole of we will have 2 moles of
Then from 0.001096 mol of :
of
According to reaction, 1 mole of reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of
of cerium (IV) ions.
Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L
1 mL = 0.001 L
![[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M](https://tex.z-dn.net/?f=%5BCe%5E%7B4%2B%7D%5D%3D%5Cfrac%7B0.004384%20mol%7D%7B0.02147%20L%7D%3D0.2042%20M)
0.2042 M is the original concentration of (aq) in the titrating solution.
Definitely D makes the most since
