The electrons and the nuclei will settle into positions that minimize repulsion and maximize attraction.
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)
One of the causes of polarity is the unequal forces of the dipole forces due to differences in electronegativity. The more electronegative molecule tend to attract electrons, hence the polarity arrows point towards it. The electronegativity of Phosphorus and Fluorine is 2.19 and 3.98, respectively. Thus, it points outward towards the fluorine atoms.
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
