39.20 ml of 0.5000 m agno3 is added to 270.00 ml
ofwater which contains 5.832 g k2cro4. a
redprecipitate of ag2cro4 forms. what is
theconcentration, in mol/l, of
unprecipitatedcro42-? be sure to enter
the correct numberof significant figures. assume
ag2cro4is completely insoluble.

concentration of CrO4²⁻ ions in the final solution = 6.53 × 10⁻⁵ mol /L

Explanation:

First we calculate the number of moles of AgNO₃:

number of moles = molar concentration × volume

number of moles = 0.5 × 39.20 = 19.6 mmoles = 0,0196 moles AgNO₃

Then we calculate the number of moles of K₂CrO₄:

number of moles = mass / molar weight

number of moles = 5.832 / 194 = 0.03 moles K₂CrO₄

The chemical reaction will look like this:

2 AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2 KNO₃

Now we devise the following reasoning:

if          2 moles of AgNO₃ are reacting with 1 mole of K₂CrO₄

then    0,0196  moles of AgNO₃ are reacting with X moles of K₂CrO₄

X = (0.0196 × 1) / 2 = 0.0098 moles of K₂CrO₄

now the the we calculate the amount of unreacted K₂CrO₄:

unreacted K₂CrO₄ = 0.03 - 0.0098 = 0.0202 moles

now the molar concentration of CrO4²⁻ ions:

molar concentration = number of moles / solution volume (L)

molar concentration = 0.0202 / (39.20 + 270) = 6.53 × 10⁻⁵ mol /L

h let me think

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