Propane torch is lit inside a hot air balloon during pre-flight preparation because the heat from the touch is needed to heat the cold air inside the balloon, so that the air will expand and become less dense and rise, thus providing a lift for the balloon. This is line with charle's law, which states that, the volume of a fixed mass of ideal gas is directly proportional to the absolute temperature. This law implies that, as the temperature of the air inside the balloon increase, the volume of the balloon also increases.
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
This is a type of metathesis reaction, also referred to as double-displacement reactions. In this reaction, the solvent and electrolyte is water, and they are driven by the formation of the non-electrolytic product. Therefore, the driving force behind the neutralization reaction between HCl and NaOH is the formation of sodium chloride, NaCl.
Answer:
= 12 mL H202
Explanation:
Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution.
Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;
we have;
(3.0 mL/ 100 mL) × 400 mL
= 12 mL H202
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:
ΔU=-369.2 kJ/mol.
