Actually, that does not happen until the protostar becomes a star when nuclear ignition starts and is maintained. It takes awhile for new star to go through its T-Tauri stage and settle down on the main sequence.
<span>A STAR does not reach hydrostatic equilibrium until it on the main sequence. Otherwise, it would remain a brown dwarf with not enough mass to to maintain nuclear fusion for more than 3,000 to 10,00 years. </span>
Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
<em>That means, 1 mole of hydroxide reacts per mole of acid</em>
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
<h3>27.6mL of LiOH 0.250M</h3>
Answer:
In covalent bonding, the octet rule is important because sharing electrons gives both atoms a full valence shell. As a result, each atom can consider the shared electrons to be part of its own valence shell.
np :)
Answer:
HCO₂/H₂O is not the acid-base conjugate pair.
Explanation:
<em>Acid and conjugate base pairs differ by an H+ ion.</em>
Neither HCO₂ nor H₂O has lost or gained protons.
The conjugate acid of H₂O is H₃O⁺
The conjugate base of HCO₃⁻ is CO₃²⁻
[A conjugate acid has one more H⁺ than its base]
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
