Answer:
11.58 L of N₂
Explanation:
We'll begin by calculating the number of mole in 37.2 g of magnesium. This can be obtained as follow:
Mass of Mg = 37.2 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 37.2 / 24
Mole of Mg = 1.55 moles
Next, we shall write the balanced equation for the reaction. This is illustrated below:
3Mg + N₂ —> Mg₃N₂
From the balanced equation above,
3 moles of Mg reacted with 1 mole of N₂.
Therefore, 1.55 moles of Mg will react with = (1.55 × 1)/3 = 0.517 mole of N₂
Thus, 0.517 mole of N₂ is need for the reaction.
Finally, we shall determine the volume of N₂ needed for the reaction as follow:
Recall:
1 mole of a gas occupies 22.4 L at STP.
1 mole of N₂ occupied 22.4 L at STP.
Therefore, 0.517 mole of N₂ will occupy = 0.517 × 22.4 = 11.58 L at STP
Thus, 11.58 L of N₂ is needed for the reaction.
Answer:
the answer should 126 859.2 m2
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31, 
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of 
= 37
So 37 mmol of acetate need to add to this solution.
