<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %
<u>Explanation:</u>
The given chemical formula of lead (IV) carbonate is
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:
Mass of lead = (1 × 207.2) = 207.2 g
Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g
Putting values in above equation, we get:
Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %
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:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
You know [A]₀, k, and t, thus you can calculate [A].