Answer:
964ug
Explanation:
The problem here involves converting from one unit to another.
We are to convert from ounces to micrograms.
1ug = 1 x 10⁻⁶g
1oz = 28.35g
So we first convert to grams from oz then take to ug:
Solving:
1oz = 28.35g
3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g
So;
1 x 10⁻⁶g = 1ug
9.64 x 10⁻⁴g will give = 9.64 x 10²ug or 964ug
<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)
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:
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