The correct option is B. The alkali metals combine directly with halogens forming high melting point crystalline solids that have high negative enthalpies of formation. The entalphy of formation for lithium bromide is the highest and the values decrease down the group, thus, lithium is the most stable while cesium is the least stable.
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Answer:
2 KOH → H₂O + K₂O
Explanation:
When KOH descomposes in potassium oxide, K₂O and water, H₂O.
The unbalanced equation is:
KOH → H₂O + K₂O
As you can see, 2 hydrogen, potassium and oxygen atoms are produced, to balance this equation we can write:
2 KOH → H₂O + K₂O
This is the balanced equation because 2 atoms of K, H and O are in reactants and products
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.
We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
It would be the Aqueous solution