0.965g of iodine was shaken with a mixture of 50cm3 trichloromethane and 50cm3 of water in the separating funnel until equilibri
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Answer:
(a) See below
(b) 103.935 °F; 102.235 °F
Explanation:
The equation relating the temperature to time is
1. Calculate the thermometer readings after 0.5 min and 1 min
(a) After 0.5 min
(b) After 1 min
2. Calculate the thermometer reading after 2.0 min
T₀ =106.321 °F
ΔT = 100 - 106.321 °F = -6.321 °F
t = t - 1, because the cooling starts 1 min late
3. Plot the temperature readings as a function of time.
The graphs are shown below.
The question is incomplete. There's missing the image, which is shown below.
Answer:
Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L
Explanation:
The reaction between hydrogen gas and oxygen gas to form water is:
2H₂(g) + O₂(g) → 2H₂O(g)
So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.
All the images are at the same temperature and pressure, so, by the ideal gas law:
PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:
P*12 = 8*RT
RT/P = 12/8 = 1.5
Thus, for O₂:
PV= nRT
V = n*(RT/P)
V = 4*1.5 = 6 L
For the mixture:
V = 12*1.5 = 18 L
For H₂O:
V = 8*1.5 = 12 L
The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L
