Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
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:
-1273.3
Explanation:
Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.
enthalpy of formation Of CO2 = 2 X -393.5 = -787
enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4
enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules
Answer:
The mass of the solute and the volume of the solution.
Explanation:
Hello,
In this case, given the formula of molarity:

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.
Best regards.