A syringe containing 1.63 mL of oxygen gas is cooled from 91.0 ∘C to 0.9 ∘C. What is the final volume Vf of oxygen gas? (Assume that the pressure is constant.) Vf = nothing mL Request Answer Provide Feedback

Vf = 1.22 mL

Explanation:

If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.

Let's convert the values:

91°C + 273 = 364K

0.9°C + 273 = 273.9K

Volume decreases if the temperature is decreases

Volume increases if the T° increases

V₁ / T₁  = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K

V₂ = (1.63mL /364K) . 273.9K → 1.22 mL

The final volume of the oxygen gas is 1.23 mL

Explanation:

Step 1: Data given

Volume of the syringe = 1.63 mL = 0.00163 L

Initial temperature = 91.0°C = 364.15 K

Final temperature = 0.9 °C = 274.05 K

Step 2: Calculate the final volume

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.00163 L

⇒with T1 = the initial temperature = 364.15 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final temperature = 274.05 K

0.00163/364.15 = V2 / 274.05

V2 = 0.00123 L = 1.23 mL

The final volume of the oxygen gas is 1.23 mL

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

2b2t

explanation:

2b2t