Answer:

Explanation:
Hello,
In this case, by knowing the given reference reactions, one could rearrange them as follows:


Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

Consequently, the equilibrium constant is computed as:
![Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%20%2A%20%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%5E2%7D%20%3DKp_2%2AKp_3%3D4.35x10%5E%7B18%7D%2A7.056x10%5E%7B-13%7D%3D3.07x10%5E6)
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Answer:Rubidium and Strontium
Explanation:
I don’t know what to do with the answer I don’t know how you feel sorry I didn’t
The scientist's results is that at a temperature of 35<span>°C, the solubility of the substance in water is 146.2 grams in 200 grams of water. There isn't really a different method to determine the solubility of a substance in water. Another procedure could be that a lesser amount of the substance is used and the water required to dissolve it is determined. The solubility of the substance based on the two procedures can then be compared.</span>