D, the more liquid there is, the less the temperature will be affected
Explanation:
cells, dna, chromosomes, body systems
The balanced chemical reaction is given as follows:
<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:
</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
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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