2hgo—> 2hg+o2. i start with 46.8 g hgo. what is my theoretical yield of o2?

b. i actually collect 2.30 g of o2. what is my percent yield of o2?

Theoretical yield = 3.52 g

Percent yield =65.34%

Explanation:

Given data:

Mass of HgO = 46.8 g

Theoretical yield of O₂ = ?

Percent yield of O₂ = ?

Actual yield of O₂ =  2.30 g

Solution:

Chemical equation:

2HgO  →   2Hg + O₂

Number of moles of HgO = mass/ molar mass

Number of moles of HgO = 46.8 g / 216.6 g/mol

Number of moles of HgO = 0.22 mol

Now we will compare the moles of HgO with oxygen.

                         HgO             :             O₂

                             2               :               1

                            0.22           :          1/2×0.22 = 0.11 mol

Theoretical yield:

Mass of oxygen = number of moles × molar mass

Mass of oxygen =  0.11 mol × 32 g/mol

Mass of oxygen = 3.52 g

Percent yield :

Percent yield = actual yield / theoretical yield × 100

Percent yield = 2.30 g/ 3.52 g × 100

Percent yield =65.34%

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

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