Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
BROO DEE YOU ALREADY KNOW WHAT IS STOP DA CAP, o.81
Answer:
The amount of drug left in his body at 7:00 pm is 315.7 mg.
Explanation:
First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:
Where:
λ: is the decay constant = 
: is the half-life of the drug = 3.5 h
N(t): is the quantity of the drug at time t
N₀: is the initial quantity
After 90 min and before he takes the other 200 mg pill, we have:
Now, at 7:00 pm we have:
Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).
I hope it helps you!
Answer:
Diluted concentration is 0.5M
Explanation:
Let's solve this with rules of three, although there is a formula to see it easier
In 1000 mL (1L), we have 2 moles of NaOH
In 250 mL we must have (250 . 2) / 1000 = 0.5 moles of NaOH
These moles will be also in 1 L of the final volume of the diluted solution
More easy:
1 L of solution has 0.5 moles of NaOH
Then, molarity is 0.5 M
The formula is: Concentrated M . Conc. volume = Diluted M . Diluted volume
2 M . 0.250L = 1L . Diluted M
0.5M = Diluted M
This isn't a good question but I guess!
