We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
Answer: 6,25 moles
Explanation: mark amount of butane x.
From equation you can calculate x in a following way:
2/x = 8/25. 8x = 50. And x = 6,25
The rate constant of a reaction : 8.3 x 10⁻⁴
<h3>Further explanation</h3>
Given
rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1
Required
the rate constant
Solution
For aA + bB ⇒ C + D
Reaction rate can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
the rate constant : k =
![\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20k%3D%5Cdfrac%7Brate%7D%7B%5BA%5D%5Em%5BB%5D%5En%7D%5C%5C%5C%5Ck%3D%5Cdfrac%7B1.10%5E%7B-2%7D%7D%7B2%5E2%5Ctimes%203%5E1%7D%5C%5C%5C%5Ck%3D8.3%5Ctimes%2010%5E%7B-4%7D)
I am going to have to say C. luster if not then B. Cleavage
