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 correct answer is:

a. formation of precipitation

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.