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.
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Answer:
see explanation...
Explanation:
Mg⁺²-24 Co⁺³-60 Clˉ-35
Protons (p⁺) 12 27 17
Neutrons (n⁰) 12 33 18
Electrons (eˉ) 10 24 18
(c) (b) (a)
12/2 : 12/2 : 10/2 27/3 : 33/3 : 24/3 #n⁰ = 18
6 : 6 : 5 9 : 11 : 8 #eˉ = 18
Answer:
Yes
Explanation:
Is this a question or what?
The percentage yield of the new production technique is 82.8%
<h3>What is the percentage yield?</h3>
Production is the procedure by which finished products are obtained form the raw materials. The production process involves the passing of raw materials through a certain procedure that involves the use of certain machines and equipment to give us the required products.
We are told in the question that there are three shifts;
Shift 1 produces 4562 grams
Shift 2 produces 5783 grams
Shift 3 produces 5247 grams
Average production from the three shifts = 4562 grams + 5783 grams + 5247 grams/3 = 5197 grams
The theoretical average yield is = 7000 grams + 7000 grams + 7000 grams/3 = 7000 grams
Now the percentage yield = actual yield/ theoretical yield * 100/1
percentage yield = 5197 grams/7000 grams * 100/1
percentage yield = 82.8%
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Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:

Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:


Therefore, the maximum amount of nickel(II) cyanide is 