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Chemistry

1 answer:

Anastasy [175]4 years ago

8 0

Been awhile since I done these but i believe it looks like this

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Question 4 to make a 0.500 m solution, one could take 0.500 moles of solute and add 1.00 l of solvent. enough solvent to make 1.

stealth61 [152]

The answer is enough solvent to make 1.00 L of solution. Since molarity is the number of moles of solute in one liter of solution, adding 0.500 mole solute to one liter solvent might not result to a solution with one liter total volume. Less than one liter solvent is first added to dissolve 0.500 mole solute and then the solution is carefully filled with more solvent until the solution reaches to one liter total volume. Hence, the resulting solution is a 0.500M concentration.

7 0

3 years ago

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How many grams of Ag2S2O3 form

Art [367]

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

<h3>Mass of Ag₂S₂O₃ formed</h3>

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?

$mass of Ag_{2} S_{2} O_{3} =\frac{125 grams of AgBrx327.74 grams of Ag_{2} S_{2} O_{3}}{375.54 grams of AgBr}$