Explanation:
Formula to calculate hybridization is as follows.
Hybridization =
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
So, hybridization of
is as follows.
Hybridization =
=
= 2
Hybridization of
is sp. Therefore,
is a linear molecule. There will be only two electron groups through which Be is attached.
Similarly, hybridization of
is calculated as follows.
Hybridization =
=
= 5
Therefore, hybridization of
is
is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.
Thus, we can conclude that out of the given options
is the correct examples of linear molecules for five electron groups.
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)

Mass of Ca²⁺ in 2.00 L urine sample is:

= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:

= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
C. quadruples the rate
<h3>Further explanation</h3>
Given
The rate law :
R=k[A]²
Required
The rate
Solution
There are several factors that influence reaction kinetics :
- 1. Concentration
- 2. Surface area
- 3. Temperature
- 4. Catalyst
- 5. Pressure
- 6. Stirring
The rate is proportional to the concentration.
If the concentration increased, the reaction rate will increase
The reaction is second-order overall(The exponent is 2)
The concentration of A is doubled, the reaction rate will increase :
r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²
<em>The reaction rate will quadruple.</em>