Hi!
The Samurai is what I would attribute to the memory of Ancient<span> Japan.</span>
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
Answer: On increasing temperature at which adsorption is carried out decreases the extent of physisorption.
Explanation:
An adsorption where molecules of the adsorbate are placed or held on the surface of adsorbent by Vander waals forces is called physisorption.
There is basically physical bonding between the molecules of gas to the surface of a solid or liquid.
Physisorption is reversible in nature and occurs at low temperatures.
It is not specific in nature which means that all gases are adsorbed on the surface of every solid substance to some extent.
Thus, we can conclude that on increasing temperature at which adsorption is carried out decreases the extent of physisorption.
Answer is: molarity of hydrofluoric solution is 0.09 M.
Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
c(KOH) = 0.122 M.
V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
Is there choices? Cuz if there is i Need them
