<h2>2 moles of NH³ </h2>
<h2>MARK ME BRAINLIST</h2>
M₁=6.584 g
m₂=4,194 g
m(H₂O)=m₁-m₂
w(H₂O)=m(H₂O)/m₁
w(H₂O)=(m₁-m₂)/m₁
w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)
the percentage by mass of water in the hydrate 36.30
14. is a and i think 15. is b
Answer: A, C, E
Explanation: PLATO. all testable questions.
Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles = 
=
= 0.001245 moles
Concentration of KHP (C1) in litres = n/v
=
= 0.062 mol/L
We know that:
=
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>