Answer:
Diameter He = 0,1 nm.
Explanation:
Km to nm:
⇒ Diameter He = 1.0 E-13 Km * ( 1000 m / Km ) * ( 1 E9 nm / m )
⇒ Diameter He = 0.1 nm
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>
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
KOH is a compound containing two ions, K+ and OH-.
<span>The polyatomic ion present is OH- which is called hydroxide. </span>
<span>The compound is named potassium hydroxide.</span>
By studying and learning by using your brain to accomplish new goals. :)
(if that makes sense)
Hope this helps :3
