They do this so that when you drink it or use it the contaminates that used to be in it dont make you sick
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Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = 
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)


According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
Answer:
The answer to your question is below
Explanation:
Data
Volume = 1000 ml
Concentration = 2M
molecule = NaCl
Process
1.- Calculate the number of moles of NaCl
Molarity = moles/Volume
-Solve for volume
moles = Molarity x Volume (liters)
-Substitution
moles = 2 x 1
-Result
moles = 2
2.- Determine the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5 g
3.- Calculate the mass of NaCl to prepare the solution
58.5 g ----------------- 1mol
x ----------------- 2 moles
x = (2 x 58.5) / 1
x = 117g
4.- Weight 117 g of NaCl, place them in a volumetric flask (1 l), and add enough water to prepare the solution.
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