All elements in their standard states (oxygen<span> gas, solid carbon in the form of graphite, etc.) have a standard </span>enthalpy of formation<span> of </span>zero<span>, as there is no change involved in their </span>formation<span>.</span>
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer: 84.56L
Explanation:
Initial volume of gas V1 = 100L
Initial temperature T1 = 135°C
Convert temperature in Celsius to Kelvin
( 135°C + 273 = 408K)
Final temperature T2 = 72°C
( 72°C + 273= 345K)
Final volume V2 = ?
According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.
Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2
100L/408K = V2/345K
To get the value of V2, cross multiply
100L x 345K = V2 x 408K
34500 = V2 x 408K
V2.= 34500/408
V2 = 84.56L
Thus, the volume of the gas becomes 84.56 liters