In the copper – silver nitrate lab copper medals and silver nitrate solution reacted to produce silver metal and copper (ii) nitrate in solution. a student placed a copper wire with a mass of 2.93 g in the reaction test tube. the silver nitrate solution contained 1.41 g of silver nitrate. he obtained .87 g of silver metal. calculate the percent yield of silver.

Percent Yield = 97.75 %

Explanation:

1 MOLE = It is equal to the molar mass of the substance

1 mole of Cu = 63.54 g (Molar Mass of Cu = 63.54 g/mole)

1 mole of AgNO3 = 170 g (Molar Mass of AgNO3 = 170 g/mol)

Given Mass of AgNO3 = 1.41 g

Given Mass of Cu = 2.93 g

Second step : Find the limiting Reagent (which is in less amount)

Balanced Chemical equation :

This means

1 mole of Cu will react with  = 2 mole of AgNO3

63.54 g of Cu  reacts with = 2 x 170 g of AgNO3

1 g  of Cu  reacts with = (2 x 170)/63.54 of AgNO3

= 5.35 g of AgNO3

2.93 g should reacts with = 2.93 x 5.35 = 15.67 g of AgNO3

Available AgNO3 = 1.41 g

So , AgNO3 is less than required = limiting reagent

Now the reaction occur 1.41 g of AgNO3

Now, Limiting reagent will decide How much Silver(Ag) Metal will form

2 mole of AgNO3 will produce = 2 mole of Ag

1 mole of AgNO3 will produce = 1 mole of Ag

170 g  AgNO3 will produce = 107.86 mole of Ag(Molar mass of Ag = 107.86)

1 g AgNO3 will produce =

1.41 g of AgNO3 will produce =

= 0.89 g

Actual yield = 0.87 g

Theoritical yield = 0.89 g

Percent Yield = 97.75 %

im pretty sure its b

he uptake of drugs from the intestine is often strongly influenced by diffusion. for example, consider a water-insoluble steroid for treating inflammation that is solubilized in aggregates of detergent molecules, called micelles. these micelles, aggregates of 80 soap molecules, have a molecular weight of 24,000, a diameter of 2.6 nm, and a charge of -27. the positively charge ions (counterions) in this system are na".