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:
V2 = 9.58 Litres.
Explanation:
Given the following data;
Initial volume = 25 L
Initial pressure = 115 kPa
Final pressure = 300 kPa
To find the new volume V2, we would use Boyles' law.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
Substituting into the equation, we have;
V2 = 9.58 L
Therefore, the new volume is 9.58 litres.
<span>This means that has great capacity to react with other chemical elements in nature, reacting mainly with sodium, therefore, can only be findings in chemical compounds in most cases.
hope this helps!.</span>
0.01772288- Calculators do come in handy for this haha
Answer:
They get more energy, so they vibrate!
Explanation:
