Answer:
C
Explanation:
Because on the moon you are basically weightless and your mass would stay the same
Answer:
The pressure contribution from the heavy particles is 17.5 atm
Explanation:
According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.
In the simulation:
the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,
the total pressure of the mixture of 150 heavy and 50 light particles was measured to be 23.4 atm
Total pressure = partial pressure of Heavy particles + partial pressure of light particles
23.4 atm = partial pressure of Heavy particles + 5.9 atm
Partial pressure of Heavy particles = (23.4 - 5.9) atm
Partial pressure of Heavy particles = 17.5 atm
Therefore, the pressure contribution from the heavy particles is 17.5 atm
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Answer:
2AL(ClO3)3 → 2ALCl3 + 9O2
Explanation:
