The answer is A warm air rises cool air sinks
Answer:
The answer to your question is: 85.458 amu
Explanation:
data
Rubidium-85 A = 84.9118 amu abundance = 72.15%
Rubidium - 87 A = 86.9092 amu abundance = 27.85%
Atomic weight = ?
Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)
Atomic weight = 61.2538 + 24.2042
Atomic weight = 85.458 amu
This is an incomplete question, here is a complete question.
A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸
Answer : The concentration of
ions at equilibrium is, 
Explanation : Given,
Moles of
= 0.130 mol
Volume of solution = 1 L

Concentration of
= Concentration of
= 0.130 M
Concentration of
= 1.20 M

The equilibrium reaction will be:
![Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%2B6NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D)
Initial conc. 0.130 1.20 0
At eqm. x [1.20-6(0.130)] 0.130
= 0.42
The expression for equilibrium constant is:
![K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5BNi%28NH_3%29_6%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5BNH_3%5D%5E6%7D)
Now put all the given values in this expression, we get:


Thus, the concentration of
ions at equilibrium is, 
B: Convection... Because it is a Repeated action of heat just like Convection heating
Answer:
okay let me know when i can help, ill be happy too help