Answer:
The two main variables in an experiment are the independent and dependent variable. An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.
Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.
Answer:
1.204 × 10²⁴ atoms
Explanation:
According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.
If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum
2 × 6.02 × 10²3
= 12.04 × 10²³
= 1.204 × 10²⁴ atoms.
Answer:
1.204 × 10^24 formula units
Explanation:
Using the following formula to convert mass of iron (III) chloride to moles;
mole = mass/molar mass
Molar mass of FeCl3 = 56 + 35.5(3)
= 56 + 106.5
= 162.5g/mol
Mole = 325/162.5
mole = 2mol
Next, to calculate the formula units of FeCl3, we multiply the number of moles of FeCl3 by Avagadro number (6.02 × 10^23)
Formula units of FeCl3 = 2 × 6.02 × 10^23
12.04 × 10^23
= 1.204 × 10^24 formula units.
Answer:
Explanation:
The<em> half-life </em>time of a radiactive isotope (radioisotope) is a constant value, meaning that the amount of the radioisotope that decays will be (1/2) raised to the number of half-lives passed.
Naming A₀ the initial amount to the radioisotope, you can build this table to find the amount left.
Number of half-lives amount of radiosotope left
0 A₀
1 (1/2) × A₀
2 (1/2)×(1/2)×A₀ = (1/2)² × A₀
3 (1/2)³ ×A ₀
4 (1/2)⁴ × A₀
n (1/2)ⁿ × A₀
Now calculate the number of half-lives the strontium-90 sample has passed after 100 years:
- n = 100 years / 28.1 years ≈ 3.5587
Hence, the amount of strontium-90 is:
![(\frac{1}{2})^{3.5587}A_0=0.08486A_0](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B3.5587%7DA_0%3D0.08486A_0)
In percent, that is:
![(0.08486A_0/A_0).100=8.486%](https://tex.z-dn.net/?f=%280.08486A_0%2FA_0%29.100%3D8.486%25)
Rounding to two significant figures, that is 8.5%.
<u>Conclusion</u>: <em>The percent of strontium-90 left after 100 yeaers is 8.5% </em>(choice number 4).