Answer: A volume of 600 mL of 3.0 M solution can be prepared by using 100.0 mL OF 18 M .
Explanation:
Given: = ?,
,
Formula used to calculate the volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 600 mL of 3.0 M solution can be prepared by using 100.0 mL OF 18 M .
Answer:
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
Mass of Aluminum = 157.5 g
Given
17.5 g of Hydrogen
Required
Mass of Aluminum
Solution
Reaction
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
mol Hydrogen :
= mass : molar mass
= 17.5 g : 2 g/mol
= 8.75
From the equation, mol Al :
= 2/3 x mol H₂
= 2/3 x 8.75
= 5.83
Mass Al :
= 5.83 x 27 g/mol
= 157.5 g