Due to having STP conditions and given volume you can use this conversion: 1 mol=22.4L at STP. Since you have mL, convert to Liters first. 16800 mL = 16.8 L. You will need molar mass of Oxygen(O2)
16.8 L O2 x(1 mol/22.4 L) x( 31.998 g O2/1 mol) = 24.0 g O2
Alternatively, maybe you did not know 1 mol =22.4 L conversion, you can use PV=nRT will work, with P = 1 atm, R =0.08206 L(atm)/mol(k), and T=273 K, V = 16.8 L. Solve for n(moles), then multiply by molar mass of O2
Answer:
Heat, temperature, and thermal energy are related because they all work with each other.
Explanation:
First of all, everything start's off with temperature. It starts off low. But when heat is added to it, it rises and the temperature goes up. This causes thermal energy to the objects touching it. The hotter it is the faster the particles move and the more kinetic energy they have.
Answer:
176 g of CO₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O
Next, we shall determine the mass of C₂H₂ that reacted and the mass of CO₂ produced from the balanced equation.
This is illustrated below:
Molar mass of C₂H₂ = (2×12) + (2×1)
= 24 + 2 = 26 g/mol
Mass of C₂H₂ from the balanced equation = 2 × 26 = 52 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32 = 44 g/mol
Mass of CO₂ from the balanced equation = 4 × 44 = 176 g
From the calculations made above,
52 g of C₂H₂ reacted to produce 176 g of CO₂.
The Barometer measures the pressure while the <span>Psychrometer measures the temperature.</span>
Answer:
15.71g
Explanation:
The general combustion equation for all hydrocarbons is
CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O
For octane, C8H18 :
C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O
C8H18 + 50/4 O2 = 8CO2 + 9H2O
C8H18 + 25/2 O2 = 8CO2 + 9H2O
2C8H18 + 25 O2 = 16 CO2 + 18H2O (balanced)
From the balanced equation,
2 x 22.4 L of octane produced 16 [ 12 + (16 x 2)] of carbon dioxide
That is,
44.8 L of octane produced 704g of carbon dioxide
So, 1L of octane will produce 1 L x 704g/44.8 L = 15.71g of carbon dioxide
Therefore, 15.71g of carbon dioxide will be produced by the complete combustion of 1 L of octane.
