Use the balanced equation below to determine the mass produced of Ca3(PO4)2precipitate that would be produced if 35 ml aqueous 0.0025 M Ca(NO3)2reacts with excess Na3PO4? Molar Mass: 310.18 g/mol Ca3(PO4)23Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq) a.0.00905 g

b.8.75 x 10-5g

c.9.40 x 10-8g

d.0.905 g

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

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

catya that's not even a b or c