Answer:
32
Explanation:
because the number of heliem is your aswer
Hope this helps! 159.2086<span>Have a Good Day!</span>
Answer:
Average atomic mass = 51.9963 amu
Explanation:
Given data:
Abundance of Cr⁵⁰ with atomic mass= 4.34%
, 49.9460 amu
Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu
Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu
Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu
Average atomic mass = 51.9963 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100
Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100
Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100
Average atomic mass = 5199.632 / 100
Average atomic mass = 51.9963 amu
Answer:
0.144 nm
Explanation:
Silver's electronic configuration is (Kr)(4d)10(5s)1, and it has an atomic radius of 0.144 nm.
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%.
