Answer is: 0.0213M.
Reaction: N₂O₄ (g) ↔ 2NO₂ (g).
Kc = 0,211 at <span>100°C.
Kc - </span><span>equilibrium constant.
</span>c (N₂O₄) = [N₂O₄] = 0,00251M.
c (NO₂) = [NO₂] = ?
Kc = [NO₂]² / [N₂O₄]
[NO₂]² = Kc · [N₂O₄]
[NO₂] = √(Kc · [N₂O₄<span>]) </span>
[NO₂] = √(0.211 · <span>0.00215M) </span>
[NO₂] = 0.0213M.
<span>NaCl is poster-compound for ionic bonding. The bonds in NaCl have about 70% ionic character, making the bond highly polar. its overstatement to state that there is actual ion in NaCl with +1 and -1 charge but actual charge of Na and Cl is +1 and -1 ion, since Nacl exist as a network of highly charged particle and not discrete molecule, NaCl particle does not exhibit intermolecular forces.
Water molecule on other hand exhibit London dispersion force, keesom force, and hydrogen bonding.
The polar water molecule are attracted to the polarized Na and Cl atoms. This is what allow NaCl(s) to dissolve and ionize in water. Therefore type of attraction in NaCl is ion-dipole attraction.</span>
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).
See figure 1
I hope it helps!
Here are the options to the question:
A) The kinetic energy of gas particles will increase.
B) The kinetic energy of gas particles will decrease.
C) The kinetic energy of gas particles will remain unchanged.
D) The gas particles will have no kinetic energy.
<span>E) The kinetic energy of gas particles will be transferred to the container.
</span>
The best answer is:
E. The kinetic energy of gas particles will be transferred to the container
At temperatures near Zero<span> K that is −273.15 °C and −459.67 °F, nearly all molecular motion ceases and ΔS = </span>0<span> for any adiabatic process, where S is the entropy.</span>