if you have an isotope of the element carbon that has the mass number of 14, what other information would help you determine the
1 answer:
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Answer:
when there is a radical in the denominator, we should rationalize (mutiply the denominator and numerator by the radical) to get rid of the radical in the denominator.
<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is moles
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
- <u>For lead (II) nitrate:</u>
Molarity of lead (II) nitrate solution = 2.70 M
Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
- <u>For NaI:</u>
Molarity of NaI solution = 0.00157 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
For the given chemical reaction:
By Stoichiometry of the reaction:
2 moles of NaI reacts with 1 mole of lead (II) nitrate
So, moles of NaI will react with =
of lead (II) nitrate
As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.
Thus, NaI is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, moles of NaI will produce =
of lead (II) iodide
Hence, the moles of precipitate (lead (II) iodide) produced is moles
Explanation:
Reaction equation is as follows.
Here, 1 mole of produces 2 moles of cations.
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with to form
and
.
As,
=
=
According to the ICE table for the given reaction,
Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
x = 0.0003 M
So, x = = 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use = 0.0003 M
The reaction will be as follows.
Initial: 0.0003
Equilibrium: 0.0003 - x x x
=
=
Therefore,
As, x <<<< 0.0003. So, we can neglect x.
Therefore,
=
x =
x = =
=
= M
Thus, we can conclude that the concentration of spectator ion is M.