Hey there!
For this we can use the combined gas law:
We are only working with pressure and temperature so we can remove volume.
P₁ = 2 atm
T₁ = 27 C
P₂ = 2.2 atm
Plug these values in:
Solve for T₂.
Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.
Hope this helps!
Carbon is the only nonmetal
Answer:
![[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Explanation:
Hello there!
In this case, for the ionization of silver iodide we have:
![AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]](https://tex.z-dn.net/?f=AgI%28s%29%5Crightleftharpoons%20Ag%5E%2B%28aq%29%2BI%5E-%28aq%29%5C%5C%5C%5CKsp%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:
![[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D%5BH%5E%2B%5D%3D10%5E%7B-3.55%7D%3D2.82x10%5E%7B-4%7DM)
Now, we can set up the equilibrium expression as shown below:
Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:
![x=[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=x%3D%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Best regards!
Answer- The particles in a solid gain enough energy to overcome the bonding forces holding them firmly in place. Typically, during melting, the particles start to move about, staying close to their neighbouring particles, then move more freely.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe
<h3>Explanation</h3>
Method One: Refer to electron transfers.
Oxidation states:
- Na: from 0 to +1; loses one electron.
- Fe: from +3 to 0; gains three electrons.
Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.
6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe
- There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
- There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Method Two: Atoms conserve.
Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.
? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe
There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.
? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.
<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
