How is an ionic formed between two atoms
2 answers:
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Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- AgBr: 2 moles
- Na₂S₂O₃: 1 mole
- Ag₂S₂O₃: 1 mole
- NaBr: 2 moles
The molar mass of the compounds is:
- AgBr: 187.77 g/mole
- Na₂S₂O₃: 158 g/mole
- Ag₂S₂O₃: 327.74 g/mole
- NaBr: 102.9 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- AgBr: 2 moles ×187.77 g/mole= 375.54 grams
- Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
- Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
- NaBr: 2 moles ×102.9 g/mole= 205.8 grams
The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?
<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>
Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
Learn more about the reaction stoichiometry:
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