Chemistry, 10.07.2019 13:00 xezthekingx4431

The label on a bottle of "organic" liquid fertilizer concentrate states that it contains 9.20 grams of phosphate per 100.0 ml and that 16.0 fluid ounces should be diluted with water to make 32.0 gallons of fertilizer to be applied to growing plants. what version of the dilution equation can be used to calculate the final phosphate concentration? express your answer in terms of vi, ci, and vf.

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Answer from: mccdp55

The final concentration of phosphate in the solution has been 0.36 g/L.

The dilution of the compound has been given as the formation of the working standard from the stock solution.

Dilution of Phosphate

The volume of the phosphate to be diluted has been 16.0 fluid ounces. The volume, vi has been given in mL as follows:

$\rm 1\;fl\;oz=29.57\;mL\\16\;fl\;oz=29.57\;\times\;16\;mL\\16\;fl\;oz=473\;mL$

The volume of the fertilizer vi containing phosphate has bee 473 mL.

The mass of phosphate in 100 0 mL has been 9.20 grams.

The mass of phosphate in 473 mL has been given as:

$\rm 100\;mL=9.2\;g\\473\;mL=\dfrac{9.2}{100}\;\times\;473\;g\\473\;mL=43.516\;g$

The mass of phosphate in 16.0 fluid ounce has been 43.516 g.

The concentration of phosphate in fertilizer, ci has been given as:

$ci=\dfrac{\text {mass}}{\text {volume}}\\ci=\dfrac{43..5\;\text g}{473\;\text {mL}} \\ci=0.092\;\rm g/mL\\\textit{ci}=92\;g/L$

The ci of the solution has been 92 g/L.

The final volume of the solution, vf has been given as 32 gallons. The volume of solution has been:

$\rm 1\;gallon=3.785\;L\\32\;gallon=3.785\;\times\;32\;L\\32\;gallon=121.1\;L$

The vf of the solution has been 121.1 L.

The final concentration of the dilution has been:

c_iv_i=c_fv_f

Substituting the values in the equation:

$\rm 92\;\times\;0.473=c_f\;\times\;121.1\\c_f=\dfrac{92\;\times\;0.473}{121.1}\\c_f=0.36\;g/L$

The final concentration of phosphate in the solution has been 0.36 g/L.

Learn more about concentration of phosphate, here

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Answer from: mel4245

The final phosphate concentration is 0.36 g/L.

I would express the concentrations in grams per litre and then use the dilution equation.

V_i = 16 fl oz × (29.57 mL/1 fl oz) = 473 mL

Mass of phosphate = 473 mL × (9.20 g/100.0 mL) = 43.5 g

c_i = 43.5 g/473 mL = 0.0920 g/mL = 92.0 g/L

V_f = 32.0 gal × (3.785 L/1 gal) = 121.1 L

c_iV_i = c_f V_f

c_f = c_i × V_i/V_f = 92.0 g/L × (0.473 L/121.1 L) = 0.36 g/L

The final phosphate concentration is 0.36 g/L.

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

Answer from: Quest

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.

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