Answer:
Rubidium-85=61.2
Rubidium-87=24.36
Atomic Mass=85.56 amu
Explanation:
To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.
<u>Rubidium-85 </u>
This isotope has an abundance of 72%.
Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.
- 72/100= 0.72 or 72.0 --> 7.2 ---> 0.72
Multiply the mass of the isotope, which is 85, by the abundance as a decimal.
- mass * decimal abundance= 85* 0.72= 61.2
Rubidium-85=61.2
<u>Rubidium-87</u>
This isotope has an abundance of 28%.
Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.
- 28/100= 0.28 or 28.0 --> 2.8 ---> 0.28
Multiply the mass of the isotope, which is 87, by the abundance as a decimal.
- mass * decimal abundance= 87* 0.28= 24.36
Rubidium-87=24.36
<u>Atomic Mass of Rubidium:</u>
Add the two numbers together.
- Rb-85 (61.2) and Rb-87 (24.36)
Answer:
The symbol of each element is, Ne, Na, Mg, and Al.
Explanation:
Below is the list of elements that has an atomic mass of less than 19.3 u.
The atomic mass of Neon is 20.1797 u and the atomic number is 10.
The atomic mass of Sodium is 22.989769 u and the atomic number is 11.
The atomic mass of Magnesium is 24.305 u and the atomic number is 12.
The atomic mass of Aluminium is 26.981539 u and the atomic number is 13.
Here, the symbol of each element is, Ne, Na, Mg, and Al.
Answer:
Li(s) + H20(l) --> LiOH(aq) + H(g)
Explanation:
it is already balanced
The equation is
W = C/F
W= 3.00 x 10^8 m/sec
——————————
6.165 x 10^14 Hz
W= 4.87 x 10^-7 m
Energy is
E=hF
E= (6.626 x 10^-34 Jxsec )(6.165 x 10^14 Hz)
E= 4.085 x 10^-19 J
