D
Because of the OH substituent
<span>do not move through the earth
L waves are surface waves.
</span>
17.) C 4-methylhex-1-ene
5.) D oct-3-ol-5-en
Answer:
0.005404 M
Explanation:
![Pb^{2+}(aq) + Na_{2}CO_{3}(aq) ---> PbCO_{3}(s) + 2Na^{+}(aq)](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%28aq%29%20%2B%20Na_%7B2%7DCO_%7B3%7D%28aq%29%20---%3E%20PbCO_%7B3%7D%28s%29%20%2B%202Na%5E%7B%2B%7D%28aq%29)
Since you added an excess of sodium carbonate you warrantied that all the
in the sample reacted with it. So we can say that the insoluble lead (II) carbonate
contains all the
ions in the original sample.
The moles of
are:
![moles-of-PbCO_{3}=\frac{mass-of-PbCO_{3}}{Molecular-weight-of- PbCO_{3}}=\frac{0.1443g}{267\frac{g}{mol}}=0.00054 mol](https://tex.z-dn.net/?f=moles-of-PbCO_%7B3%7D%3D%5Cfrac%7Bmass-of-PbCO_%7B3%7D%7D%7BMolecular-weight-of-%20PbCO_%7B3%7D%7D%3D%5Cfrac%7B0.1443g%7D%7B267%5Cfrac%7Bg%7D%7Bmol%7D%7D%3D0.00054%20mol)
One mol of
is required to form one mol of
. So, the stoichiometric relationship between them is 1:1.
Knowing this, 0.00054 is also the number of moles of
in the original sample.
So, the concentration of
in the original sample is:
![M = \frac{mol-of-Pb^{+2}}{volume-wastewater-(liters)}=\frac{0.00054}{0.1L}=0.005404 M](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bmol-of-Pb%5E%7B%2B2%7D%7D%7Bvolume-wastewater-%28liters%29%7D%3D%5Cfrac%7B0.00054%7D%7B0.1L%7D%3D0.005404%20M)
Answer:
0.59 moles
Explanation:
Data Given
oxygen = 35 g
Moles of SeO₃ = ?
Reaction Given
2Se + 3O₂ ------> 2SeO₃
Solution:
Step 1.
First to find grams of SeO₃
So, we know from reaction that 2 mole of selenium combine with 3 mole of oxygen and produces 2 mole of SeO₃
2Se + 3O₂ ------> 2SeO₃
2mol 3 mol 2 mol
If we represent mole in grams
Then,
Molar mass of Se = 79 g/mol
Molar mass of O = 16 g/mol
Molar mass of SeO₃ = 79 + 3(16)
Molar mass of SeO₃ = 127 g/mol
2Se + 3O₂ ------> 2SeO₃
2mol (79 g/mol) 3 mol (32 g/mol) 2 mol (127 g/mol)
2Se + 3O₂ ------> 2SeO₃
158 g 96 g 206 g
It is obvious from the reaction that 96 g of oxygen gives 206 g of SeO₃.
Now how many grams of SeO₃ will produce if 35 grams of oxygen react with excess of Selenium
Apply unity formula
96 g of O₂ ≅ 206 g of SeO₃
35 g of O₂ ≅ x g of SeO₃
By doing cross multiplication
g of SeO₃ = 206 x 35 / 96
g of SeO₃ = 75 g
Step 2.
Convert grams of SeO₃ to mole
Formula used
no. of moles = mass in grams (SeO₃) / Molar mass of SeO₃
Put values in above formula
no. of moles = 75 g / 127 g/mol
no. of moles = 0.59 mol
So,
35 g of oxygen produces 0.59 moles of SeO₃