The empirical formula of the compound is C. NiF₂.
<em>Step 1</em>. Calculate the <em>moles of each element</em>
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ni to F.
Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni
Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F
<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smallest number of moles
Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = NiF₂
Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
Explanation:
A synthesis reaction: It is defined as a kind of reaction where one and more than one reactant attached and creates an individual product.
The formation of the water is an example of a synthesis reaction because here more than one reactants combine and create a single product (water). Water formation occurs when 2 hydrogens and an oxygen share electrons through covalent bonds.
2H + O ----> H2O.
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:
K = 0.2
Explanation:
Based on the chemical dissociation of N₂O₄:
N₂O₄ ⇄ 2NO₂
The equilibrium constant, K, of the reaction is:
K = [NO₂]² / [N₂O₄]
Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M
as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.
Replacing in K:
K = [0.4M]² / [0.8M]
<h3>K = 0.2</h3>