What is the final [Na+] in a solution prepared by mixing 70.0 mL of 3.00 M Na2SO4 with 30.0 mL of 1.00 M NaCl?

4.5 M

Explanation:

70.0 ml was mixed in 3.00 M of Na2SO4

30.0 ml was mixed in 1.00 M of NACL

The first step is to convert 70 ml to liters

= 70/1000

= 0.07 liters

The formular for molarity is

moles/liters

The number of moles in Na2S04 can be calculated as follows

Let y represent the number of moles

3M= y moles/0.07

= 3×0.07

= 0.21 moles

Since Na2So4 has 2 moles of Na then the number of moles is

= 2×0.21

= 0.42 moles

Convert 30ml to liters

= 30/1000

= 0.03 liters

The number of moles in Nacl can be calculated as follows

Let y represent the number of moles

1M= y moles/0.03

= 1×0.03

= 0.03 moles

Since Nacl has 1 mole of Na then the number of moles is

= 1 × 0.03

= 0.03 moles

Therefore the final Na+ can be calculated as follows

Total moles = 0.03 moles + 0.42 moles

= 0.45 moles

Total liters= 0.07 liters + 0.03 liters

= 0.1 liters

Na+ = 0.45/0.1

= 4.5 M

Hence the final Na+ in the solution is 4.5 M

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