Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
For plato users
the answer is a. O2(l) O2(g)
hope this helps!
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
12 moles are in the 175L of SO2 gas at STP
