Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
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
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
Answer : The volume of the cube submerged in the liquid is, 29.8 mL
Explanation :
First we have to determine the mass of ice.
Formula used :

Given:
Density of ice = 
Volume of ice = 45.0 mL


The cube will float when 40.5 g of liquid is displaced.
Now we have to determine the volume of the cube is submerged in the liquid.



Thus, the volume of the cube submerged in the liquid is, 29.8 mL
Answer:
The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Explanation:
Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg
The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)
where u = initial rms speed of the gas = 850 m/s
Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J
Then, 3600 J is added to the gas,
New kinetic energy of the gas = 4335 + 3600 = 7935 J
New kinetic energy of Helium atoms = (1/2)(m)(v²)
where v = final rms speed of the gas = ?
7935 = (1/2)(0.012)(v²)
v² = (7935×2)/0.012
v² = 1,322,500
v = 1150 m/s
Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Hope this Helps!!!