The term amphoteric describes a substance that can act as both an acid and a base.
Because you need to know what you are looking for before actually trying something so you can prevent any accidents by doing stuff at random
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
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