If O2O2 is collected over water at 40.0 ∘C∘C and a total pressure of 755 mmHgmmHg , what volume of gas will be collected for the complete reaction of 23.87 gg of NiONiO? The vapor pressure of water at 40.0 °C is 55.4 mmHg. The molar mass of NiO is 74.69 g/mo

The correct answer is 4.46 L.

Explanation:

Based on the given information, the mass of NiO given is 23.87 grams. The no. of moles can be determined by using the formula, n = weight/molecular mass. The molecular mass of NiO is 74.69 gram per mole. Now the moles of NiO will be,

Moles of NiO = 23.87 g/74.69 g/mol = 0.3196 moles

Now the moles of O2 will be = 1/2 * moles of NiO

= 1/2 * 0.3196 = 0.160 moles

The total pressure given in the question is 755 mmHg, and the vapor pressure mentioned is 55.4 mmHg. So, the pressure of O2 will be,

= Total pressure - Vapor pressure

= 755 mmHg - 55.4 mmHg

= 699.6 mmHg

= 699.6 * (1 atm / 760 mmHg)

= 0.9205 atm

Now the volume of the oxygen gas collected can be calculated by using the ideal gas law equation, that is, PV = nRT

0.9205 atm . V = 0.160 moles * 0.08206 L-atm/mol-K * 313 K (273 + 40 K)

V = 0.160 moles * 0.08206 L-atm/mol-K * 313 K / 0.9205 atm

V = 4.46 L

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