Answer:
280 g Al₂O₃
Explanation:
To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).
Molar Mass (Al₂O₃): 101.96 g/mol
2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole
Answer:
Answer to you need to make a 6.00 x 10-4 M KSCN solution starting with a 2.00 x 10-3 M KSCN solution. You will be making up the so. ... X 10-3 M KSCN Solution. You Will Be Making Up The Solution In A 25 ML Volumetric Flask And Using 0.5 M HNO3 As The Diluent. What Volume Of 2.00 X 10-3 M KSCN Will You Need?
Explanation:
Kc' =Kc^1/3
=3√0.0061
=0.182716013
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
-OH
Explanation:
Alcohols generally have the structural formula OH
for example, ethanols structural formula is C2H5OH
