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Vaselesa [24]
2 years ago
11

A 0.56 M solution of AlCl₃ is determined to have a concentration of particles of 1.79 M. What is the van't Hoff factor for AlCl₃

?
Chemistry
1 answer:
Crank2 years ago
3 0

Answer:

Van't Hoff factor for AlCl₃ = 3 (Approx)

Explanation:

Given:

Number of observed particular = 1.79 M

Number of theoretical particular = 0.56 M

Find:

Van't Hoff factor for AlCl₃

Computation:

Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular

Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M

Van't Hoff factor for AlCl₃ = 3.19

Van't Hoff factor for AlCl₃ = 3 (Approx)

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How many milliliters of 0.0630 m edta are required to react with 50.0 ml of 0.0110 m cu2 ?
gizmo_the_mogwai [7]
Moles Cu+2 = M * V
                     =  0.05 L * 0.011  m
                     = 0.00055 moles

when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles

and when the Molarity of EDTa = 0.0630 M

∴ Volume of EDTA =  moles / Molarity
                 = 0.00055 / 0.0630
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4 0
3 years ago
3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid =  diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol

5 0
3 years ago
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Answer:

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  • They also chew the cud whereby they regurgitate food materials.
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