When solutions of silver nitrate, agno3, and calcium iodide, cai2, are mixed, a yellow precipitate of silver iodide is formed. calculate the mass of silver iodide that is formed if 50.00 ml of 1.00 m agno3 is combined with 30.00 ml of 1.25 m cai2.

Explanation:

We are given the amounts of two reactant solutions, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses, so, let's assemble our information in one place, with molar masses above the formulas and the

M_r:                                                                              234.77

                      2AgNO₃          +           CaI₂               ⟶ 2AgI + Ca(NO₃)₂

Solution: (50.00 mL, 1.00 M)    (30.00 mL, 1.25 M)

Step 1.  Calculate the moles of each reactant  

Moles of AgNO₃ = 50.00 mL × (1.00 mmol/1 mL)  = 50.00 mmol

    Moles of CaI₂ = 30.00 mL × (1.25 mmol/1 mL)  = 37.50 mmol

Step 2. Identify the limiting reactant  

Calculate the moles of AgI we can obtain from each reactant.

From AgNO₃:  

The molar ratio of AgI:AgNO₃ is 2:2.

Moles of AgI  = 50.00 mmol AgNO₃ × (2 mmol AgI/2 mmol AgNO₃)

= 50.00 mmol AgI  

From CaI₂:

The molar ratio of AgI:CaI₂ is 2:1.  

Moles of CaI₂ = 37.50 mmol × (2 mmol AgI/1 mmol CaI₂) = 75.00 mmol AgI  

AgNO₃ is the limiting reactant because it gives the smaller amount of AgI.  

Step 3. Calculate the mass of AgI.

Mass = 50.00 mmol AgI × (234.77 mg AgI /1 mmol AgI)

= 11 700 mg AgI = 11.7 g AgI

The mass of silver iodide formed is .

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current