Answer:
It covers changes to the position of equilibrium if you change concentration, pressure or temperature. ... If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change
Explanation:
Answer:
How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... be used to produce 1.99 grams of water. 1.99 mg H2O X. 1mol H2O. 18.0g X ... c. If the reaction produces 5.3 mg of carbon dioxide how many grams of water ... X. 25mol O2. 2mol C8H18. X. 32.0g O2. 1mol O2. = 4.80 x 103g O2. Answer ...
Explanation:
Answer:
The answer to your question is V = 0.32 L
Explanation:
Data
Volume of NH₃ = ?
P = 3.2 atm
T = 23°C
mass of CaH₂ = 2.65 g
Balanced chemical reaction
6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂
Process
1.- Convert the mass of CaH₂ to moles
-Calculate the molar mass of CaH₂
CaH₂ = 40 + 2 = 42 g
42 g ------------------ 1 mol
2.65 g -------------- x
x = (2.65 x 1)/42
x = 0.063 moles
2.- Calculate the moles of NH₃
2 moles of NH₃ --------------- 3 moles of CaH₂
x --------------- 0.063 moles
x = (0.063 x 2) / 3
x = 0.042 moles of NH₃
3.- Convert the °C to °K
Temperature = 23°C + 273
= 296°K
4.- Calculate the volume of NH₃
-Use the ideal gas law
PV = nRT
-Solve for V
V = nRT / P
-Substitution
V = (0.042)(0.082)(296) / 3.2
-Simplification
V = 1.019 / 3.2
-Result
V = 0.32 L
Answer: when the temperature is increased, the number of collisions per second increases.
Explanation:
the rate of collisions and the temperature is directly proportional. If the energy of the gas particles is boosted by using the temperature, the chances of the particles bumping into each other due to the high energy increases, thus increasing the number of collisions. This also increases the rate of reaction. Thus when temperature is increased the number of collisions also increases.