GEORGEOCTOBER 18, 2012Floor Wax is Harmful to the Environment
Floor wax is applied on floor surfaces to make it scuff-resistant, water-resistant, slip resistant and glossy. It provides a thin, protective and hard surface layer when applied to flooring. Conventional floor wax has five main ingredients and each one of them has detrimental effects on the environment, not to mention the chemical waste created by the continuous upkeep required. The cumulative effects of these ingredients on the environmental render their harmful actions more potent and difficult to reverse.
Energy of reactants is higher than the products , I am not sure through
hot air rises up in the balloon and lifts the balloon up
Answer:
B₂
Explanation:
The limiting reactant is always a reactant. You can determine which reactant is limiting by identifying which has the smaller mole-to-mole ratio with the product. This ratio can be found via the coefficients of the balanced reaction.
4 A₂ + 3 B₂ ---> 6 AB
4 moles A₂
------------------ = mole-to-mole ratio A₂/AB
6 moles AB
3 moles B₂
------------------ = mole-to-mole ratio B₂/AB
6 moles AB
Since the mole-to-mole ratio between B₂ and AB is smaller, B₂ must be the limiting reactant.
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
