Answer:
Option D is Correct answer
The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months
To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Amount remaining (N) = 5%
Original amount (N₀) = 100%
<h3>Number of half-lives (n) =?</h3>
N₀ × 2ⁿ = N
5 × 2ⁿ = 100
2ⁿ = 100/5
2ⁿ = 20
Take the log of both side
Log 2ⁿ = log 20
nlog 2 = log 20
Divide both side by log 2
n = log 20 / log 2
<h3>n = 4.32</h3>
Thus, 4.32 half-lives gas elapsed.
Finally, we shall determine the half-life of the element. This can be obtained as follow.
Number of half-lives (n) = 4.32
Time (t) = 500 years
<h3>Half-life (t½) =? </h3>
t½ = t / n
t½ = 500 / 4.32
t½ = 115.74 years
Multiply by 12 to express in months
t½ = 115.74 × 12
<h3>t½ ≈ 1389 months </h3>
Therefore, the half-life of the radioactive element in months is approximately 1389 months
Learn more: brainly.com/question/24868345
Answer : (4) Chromatography
Explanation :
Chromatography : It is a separation technique of a mixture by passing it through a medium in which components travels at different rates.
There are many types of chromatography but this is paper chromatography.
Paper chromatography : It is used to separate the colored substances. In paper chromatography, water is the mobile phase and paper is the stationary phase. The mixture of components moves at different speeds through the stationary phase so that they can be separated.
In paper chromatography, several colors can be separated based on their solubility. The more soluble a color is, the more readily it will dissolve in mobile phase and farther it will travel.
N₂H₄ :
N = - 2
H = + 1
2 * ( -2 ) + ( 1 * 4 ) = - 4 + 4 = 0
hope this helps!
Answer:
A general guideline to determine if oxide is acidic, basic, or amphoteric is to use the periodic table. Typically, metals such as Ba form basic oxides (BaO), while nonmetals such as S form acidic oxides (SO3).
