K5O2
convert grams to moles, divide both by the smallest mole mass, multiply that until hole.
30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2
Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:
![ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )](https://tex.z-dn.net/?f=ln%5Cfrac%7Bk2%7D%7Bk1%7D%20%3D%20%5Cfrac%7B-Ea%7D%7BR%7D%20%28%5Cfrac%7B1%7D%7BT2%7D%20-%5Cfrac%7B1%7D%7BT1%7D%20%29)
Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:
![ln\frac{k2}{k1} = 4,257](https://tex.z-dn.net/?f=ln%5Cfrac%7Bk2%7D%7Bk1%7D%20%3D%204%2C257)
![\frac{k2}{k1} = 70,593](https://tex.z-dn.net/?f=%5Cfrac%7Bk2%7D%7Bk1%7D%20%3D%2070%2C593)
![{k2} = 9,53x10^3 s^{-1}](https://tex.z-dn.net/?f=%7Bk2%7D%20%3D%209%2C53x10%5E3%20s%5E%7B-1%7D)
<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
Answer:
<span>In ionic compounds, <u>Metals</u> lose their valence electrons to form positively charged Cations.
Explanation:
Metals have the ability to loose elctrons readily. For example metals of Group IA and Group IIA readily looses electrons in order to obtain Noble Gas Configuration. On the other hand Non-metals tends to gain electrons and acquire negative charge. While Ions are made when an an element gain or loose electrons. After loosing electrons element get positive charge which is called as Cation while on gaining electron it gets negative charge called as Anion.</span>