212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
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Answer:
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<span>Step one Define the problem. Step two </span>Generate alternative solutions. Step three Evaluate and select an a. Hope this helped you!<span />
Answer:
19.5g is the theoretical yield of alum
Explanation:
Based on the balanced reaction, 4 moles of sulfuric acid produce 2 moles of alum. To solve this question we need to find the moles of H2SO4. With these moles we can find the moles of alum and its mass assuming all sulfuric acid reacts producing alum.
<em>Moles Sulfuric Acid:</em>
8.3mL = 0.0083L * (9.9mol/L) = 0.08217 moles sulfuric acid
<em>Moles Alum:</em>
0.08217 moles sulfuric acid * (2mol KAl(SO4)2•12H2O / 4mol H2SO4) =
0.041085 moles KAl(SO4)2•12H2O
<em>Mass Alum -Molar mass: 474.3884 g/mol-</em>
0.041085 moles KAl(SO4)2•12H2O * (474.3884 g/mol) =
<h3>19.5g is the theoretical yield of alum</h3>
