The pressure in the flask is 3.4 atm.
<em>pV</em> = <em>nRT
</em>
<em>T</em> = (20 + 273.15) K = 293.15 K
<em>p</em> = (<em>nRT</em>)/<em>V</em> = (1.4 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 293.15 K)/10 L = 3.4 atm
2H2+O2→2H2O
We have to be aware that the number of atoms are the same on both reactant side and product side.
Oxygen and selenium both have 6 valence electrons. They're in the same group hope this helps
Answer:
a) 157.5 grams of aluminum.
b) 1 mol
c) 9 g
Explanation:
The reaction is :

As per balanced equation
a) 3 moles of hydrogen will be produced from two moles of aluminium.
The atomic mass of aluminium = 27
therefore
3X2 grams of hydrogen is produced from 2 X 27 grams of Al
1 gram of hydrogen will be produced from
g
therefore 17.5 will be produced from = 9X 17.5 = 157.5 grams of aluminum.
b) as per balanced equation three moles or six gram of hydrogen is produced from 6 moles of NaOH.
Therefore 1 g of hydrogen will be produced from =
or 1 gram will be prepared from = 1 mole
c) from balanced equation three moles are produced from two moles of Al (27X2 = 54 g).
thus from 54 grams gives 6 grams of hydrogen
1 grams will give =