Which components can be found in both comparative and experimental investigations, but are never found in descriptive investigat
2 answers:
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Answer:
about 19 or 20 g
Explanation:
To do this, is neccesary to watch a solubility curve of this compound. This is the only way that you can know how many grams are neccesary to dissolve this compound in 50 mL of water to a given temperature.
Now, if you watched the attached graph, you can see the solubility curve of many compounds in 100 g of water (or 100 mL of water). So, to know how many do you need in 50 mL, it's just the half.
So watching the curve, you can see that at 20 °C, we simply need between 35 g and 40 g. Let's just say we need 38 grams of NH4Cl to be dissolved in 100 mL of water.
So, in 50 mL, it's just the half. So, we only need 19 g or 20 g of NH4Cl at 20 °C, to dissolve this compound in water.
Answer:
a) rate law = k[HgCl₂][C₂O₄²⁻]²
b) K = 8.67 * 10⁻⁵M⁻²s⁻¹
c) rate = 2.55 * 10⁻⁵Ms⁻¹
Explanation:
a) From the given data, rate law is given as: rate = k[HgCl₂]ᵃ[C₂O₄²⁻]ᵇ
Note: Further calculations and explanations are found in the attachment below
