Answer:
- The abundance of 107Ag is 51.5%.
- The abundance of 109Ag is 48.5%.
Explanation:
The <em>average atomic mass</em> of silver can be expressed as:
107.87 = 106.90 * A1 + 108.90 * A2
Where A1 is the abundance of 107Ag and A2 of 109Ag.
Assuming those two isotopes are the only one stables, we can use the equation:
A1 + A2 = 1.0
So now we have a system of two equations with two unknowns, and what's left is algebra.
First we<u> use the second equation to express A1 in terms of A2</u>:
A1 = 1.0 - A2
We <u>replace A1 in the first equation</u>:
107.87 = 106.90 * A1 + 108.90 * A2
107.87 = 106.90 * (1.0-A2) + 108.90 * A2
107.87 = 106.90 - 106.90*A2 + 108.90*A2
107.87 = 106.90 + 2*A2
2*A2 = 0.97
A2 = 0.485
So the abundance of 109Ag is (0.485*100%) 48.5%.
We <u>use the value of A2 to calculate A1 in the second equation</u>:
A1 + A2 = 1.0
A1 + 0.485 = 1.0
A1 = 0.515
So the abundance of 107Ag is 51.5%.
Answer:
9.51 × 10⁴ kL
Explanation:
Step 1: Given data
Volume of the sample (V): 9.51 × 10⁹ cL
Step 2: Convert "V" to liters
We will use the conversion factor 1 L = 100 cL.
9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L
Step 3: Convert "V" to kL
We will use the conversion factor 1 kL = 1000 L.
9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL
9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.
Answer:
Eukarya
Explanation:
It's larger than a prokaryotic cell and it has a nucleus.
Answer:
Approximately 5.646 * 10^-17 moles
Explanation:
Avagardo's number is approximately 6.022 * 10^23 molecules. Therefore, dividing 3.4 * 10^7 by avagadro's number yields approximately 5.646 * 10^-17 moles. Hope this helps
