Answers:
A) 2040 kg/m³; B) 58 600 km
Explanation:
A) Density


<em>B) Radius</em>



![r= \sqrt [3]{ \frac{3V }{4 \pi } }](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3V%20%7D%7B4%20%5Cpi%20%7D%20%7D)
![r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3%5Ctimes%208.268%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%7B4%20%5Cpi%20%7D%20%7D%3D%20%5Csqrt%20%5B3%5D%7B%201.974%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%3D%205.82%20%5Ctimes%2010%5E%7B7%7D%20%5Ctext%7B%20m%7D%3D%5Ctext%7B58%20200%20km%7D)
1) Ca-37, with a half-life of 181.1(10) ms.
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
, 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>
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Answer:
Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Explanation:
The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.
The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.
Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.