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 new pressure, if volume and amount of gas do not change is 2.40 atm
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:

Putting values in above equation, we get:

Thus the new pressure, if volume and amount of gas do not change is 2.40 atm
Answer:The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
Explanation:
Metals are usually arranged in an order of reactivity called activity series. Metals that are high up in the series are good reducing agents with very low (very negative) reduction potentials. Metals with greater (less negative) reduction potentials are found lower in the series. In the image attached, elements were arranged according to their reducing ability. Magnesium is very electro positive hence it is a better reducing agent with a lesser standard reduction potential than lead(refer to the image for numerical values of standard reduction potentials). Hence it displaces lead from solution and the elemental lead deposits on the wire.
0.075 L * 1.0 M = 0.075 mol HCl
H2 is half of HCl by the coefficients divide 0.075 by 2 and get 0.0375 mol H2 gas