He could be blindfolded and know which was his and which was his sister's. All he would need to do is pick them both up and if they were too big then pick them up one at a time. The lumber might make it harder to tell, but this is a question about physical properties.
So there is a change in mass which for the purpose of this question should be quite different. His sister's ought to be much lighter than his. He would find it easier to pick up.
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
Answer:
A) The mass would be the same.
Explanation:
Since there is no loss of any particle to vapor during the phase change process from solid to liquid, the mass of the before and after the process will remain the same.
- In this way, the law of conservation of mass is obeyed.
- Mass is the amount of matter contained in a substance.
- Since there is no room for escape or matter loss, the mass will remain the same.
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:
According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.
From the balanced equation:
<span>1mol C3H8 requires 5mol O2 for combustion </span>
<span>Molar mass C3H8 = 44g/mol </span>
<span>8.8g C3H8 = 8.8/44 = 0.2mol C3H8 </span>
<span>This will require 5*0.2 = 1.0mol O2 </span>
<span>Molar mass O2 = 32g/mol </span>
<span>Therefore 32g of O2 required.
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