A0.200-g sample of cobalt metal reacted with hydrochloric acid according to the following balanced chemical equation: co (s) + 2 hcl (aq) ➝ cocl2 (aq) + h2 (g) the volume of hydrogen gas collected over water was 87.5 ml at 20 °c and a barometer reading of 763 mm hg. referring to example exercise 1, the adjusted volume of the h2 gas is

V H2(g) = 83.13 mL

Explanation:

Dalton's law:

∴ vapor pressure (P*)

⇒ P*H2O(20°C) = 17.5 mmHg

∴ Pbar = 763 mmHg

⇒ PH2 = 763 mmHg - 17.5 mmHg = 745.5 mmHg = 0.9809 atm

∴ Mw Co(s) = 58.933 g/mol

⇒ mol Co(s) = ( 0.200 g Co )×( mol/58.933 g ) = 3.394 E-3 mol Co

⇒ n H2(g) = ( 3.394 E-3 mol Co )( mol H2 / mol Co ) = 3.394 E-3 mol H2(g)

ideal gas:

∴ R = 0.082 atm.L/K.mol

∴ T = 20°C ≅ 293 K

the adjusted volume H2(g):

⇒ V H2(g) = ((0.082 atm.L/K.mol)(293 K)(3.394 E-3 mol)) / (0.9809 atm)

⇒ V H2(g) = 0.08313 L

⇒ V H2(g) = 83.13 mL

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

not clear