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.
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the first authors to raise the concept of
altruism</span><span>. This was later
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Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of combine with 1 mole of
Thus 0.33 mole of will combine with =
mole of
Thus is the limiting reagent as it limits the formation of product.
As 1 mole of give = 1 mole of 
Thus 0.33 moles of give =
of
Mass of 
Thus theoretical yield (g) of produced by the reaction is 9.9 grams
- No of protons=No of electrons=15
<h3>Mass no:-No of neutrons+No of protons </h3>
Now
- It has atomic no 15 hence its Phosphorus(P)
Electronic configuration:-
.
Chloride formula:-
or.
Answer:
The element with electron configuration 1s² 2s² 2p⁶ will most likely not........
Explanation:
