Answer:
i need to know the question then i can help :)
Explanation:
The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.
<h3>What are the required properties of the logarithm?</h3>
The required logarithm properties are
logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);
Where a is the base of the logarithm.
<h3>Calculation:</h3>
It is given that,
log₄(x) = 12;
On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;
So,
log₄(x) = 12 ⇒ 4¹² = x
⇒ x = (2²)¹² = 2²⁴
Then, calculating log₂(x/4):
log₂(x/4) = log₂(2²⁴/4)
= log₂(2²⁴/2²)
= log₂(2²⁴ ⁻ ²)
= log₂(2²²)
On applying the property logₐ(xⁿ) = n logₐ(x);
log₂(x/4) = 22 log₂2
We know that logₐa = 1;
So,
log₂(x/4) = 22(1)
∴ log₂(x/4) = 22.
Learn more about the properties of logarithm here:
brainly.com/question/12049968
#SPJ9
Mass of KCl= 19.57 g
<h3>Further explanation</h3>
Given
12.6 g of Oxygen
Required
mass of KCl
Solution
Reaction
2KClO3 ⇒ 2KCl + 3O2
mol O2 :
= mass : MW
= 12.6 : 32 g/mol
= 0.39375
From the equation, mol KCl :
= 2/3 x mol O2
= 2/3 x 0.39375
=0.2625
Mass KCl :
= mol x MW
= 0.2625 x 74,5513 g/mol
= 19.57 g
<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.0300 M
Initial concentration of bromine monochlorine = 0.0200 M
For the given chemical equation:
<u>Initial:</u> 0.02 0.03
<u>At eqllm:</u> 0.02-2x x 0.03+x
The expression of 
for above equation follows:
![K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr_2%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BBrCl%5D%5E2%7D)
We are given:
Putting values in above equation, we get:
Neglecting the value of x = -0.96 because, concentration cannot be negative
So, equilibrium concentration of bromine gas = x = 0.00135 M
Hence, the equilibrium concentration of bromine gas is 0.00135 M
Answer:
During deposition, a gas changes directly into a solid
Explanation:
