Answer:
Yttrium barium copper oxide (YBCO) is a family of crystalline chemical compounds, famous for displaying high-temperature superconductivity.It includes the first material ever discovered to become superconducting above the boiling point of liquid nitrogen (77 K) at about 92 K.Many YBCO compounds have the general formula Y Ba 2 Cu 3 O 7−x (also known as Y123),
Explanation:
i hope this makes sense and helps man
Answer:
We would need 10 mL of the concentrated CaCl₂ stock solution, and 30 mL of water.
Explanation:
To solve the question asked we can use the C₁V₁=C₂V₂ equation, where:
We <u>solve for V₁</u>:
We would need 10 mL of the concentrated CaCl₂ stock solution, and (40-10) 30 mL of water.
Time such as minutes, days, and months will always be a(n)....... INDEPENDENT variable
Covalent bonds are formed between atoms which have
<span>- unsatisfied valency </span>
<span>- no inert gas electronic configuration </span>
<span>- These are directional bonds </span>
<span>- formed by sharing of electrons </span>
<span>Intermolecular forces </span>
<span>- much weaker than covalent bond </span>
<span>- These are not directional (except Hydrogen bonds) </span>
<span>- These are more electrostatic in nature </span>
<span>- exist between stable molecules </span>
<span>- can be Hydrogen bonding, dipole-dipole and induced dipole-induced dipole </span>
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
