The reaction between C2H2 and O2 is as follows:
2C2H2 + 5O2 = 4CO2 + 2H2O
After balancing the equation, the reaction ratio between C2H2 and O2 is 2:5.
The moles of O2 in this reaction is 84.0 mol. According to the above ratio, the moles of C2H2 needed to react completely with the O2 is 84.0mole *2/5 = 33.6 mole.
I will use [pV/T] in the state 1 = [pV/T] in the state 2.
State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K
State 2:
p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K
Application of the formula
1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter
I believe the answer is C because the first two are mechanical hazards and C mentions plug which is electrical.
Aspartame (C₁₄H₁₈N₂O₅) is a solid used as an artificial sweetener. its combustion produces carbon dioxide gas, liquid water, and nitrogen gas
C₁₄H₁₈N₂O₅ + 16O₂-----> 14CO₂ + 9H₂O + N₂.
As it can be seen from the equation, that the coefficient of nitrogen gas in the balanced equation for the reaction is 1.
So the answer here is 1 only that is coefficient of nitrogen gas in the balanced equation for the reaction is 1.
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
