Answer:
22.6 g
Explanation:
First we <u>use the PV=nRT equation</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 24 °C ⇒ 24 + 273.16 = 297.16 K
We <u>input the data</u>:
- 7.5 atm * 2.3 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 297.16 K
And <u>solve for n</u>:
Then we <u>convert 0.708 moles of oxygen gas (O₂) to grams</u>, using its <em>molar mass</em>:
- 0.708 mol * 32 g/mol = 22.6 g
The Answer Is The Third Option
The Ammonia present in 2.5 mole of Ammonia is 42.5grams.
<h3>What is a mole ?</h3>
A mole is a measuring unit in Chemistry to measure the number of atoms in certain molar mass of a substance.
It is given that 2.5 mole of Ammonia is present
Grams of Ammonia present in 2.5 mole of Ammonia = ?
Molecular weight of Ammonia = 17 grams.
1 mole of Ammonia = 17 grams of Ammonia
2.5 mole of Ammonia = 17 * 2.5
= 42.5 grams
Therefore 42.5 grams of Ammonia is present in 2.5 mole of Ammonia.
To know more about Mole
brainly.com/question/26416088
#SPJ1
B. they collided and stuck together.