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.
A. HCl:
pH= -log [H3O+]
PH=-log (0.200)
= 0.699
poH= 14-0.699
= 13.301
b. NaOH:
PoH= -log [OH-]
= -log (0.0143)
= 1.845
pH= 14-poH
= 14- 1.845
= 12.16
c. HNO3:
PH= -log[H3O+]
=-log(3.0)
= -0.4771
poH= 14-pH
= 14-9-0.4771
= 14.4771
pH= -0.4771, poH= 14.4771
d. [Ca(OH)2] = 0.0031M
[OH-]= 2X0.0031
[OH-] = 0.0062M
PoH= - log[OH-]
=-log(0.0062)
=-log(6.2x10-3)
=-(-2.21)
= 2.21
PH=14-poH
=14-2.21
=11.79
POH=2.21, PH= 11.79
Answer:
The answer is HCl → H 2 O H+ + Cl.
To answer your question, the answer is number 1. hope this help