Nitrogen=2, Hydrogen=8, Carbon=1, Oxygen=3
Hydrogen=4, Carbon=2, Oxygen=2
Iron=1, Nitrogen=2, Oxygen=6
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Hey there!:
The hydroxyl groups act to neutralize the sodium borohydride which reduces yield.
Hope that helps!
A gas does not have a specific shape or volume (so we reject option A), instead it adjusts itself to the container (which is further influenced by other forces such as gravity and temperature) and it with time, willl fill the whole container evenly, so the correct answer is:
(2) It takes the shape and the volume of any container in which it is confined
Answer:
449 mL
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 450 mL
V₂ = ?
P₁ = 101 kPa
P₂ = 104 kPa
T₁ = 17 ºC
T₂ = 25 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (17 + 273.15) K = 290.15 K
T₂ = (25 + 273.15) K = 298.15 K
Using above equation as:
Solving for V₂ , we get:
V₂ = 449 mL
