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) :
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
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I would say that honestly reporting experimental findings is an example of using good science because science is definitely about honesty in accepting experimental findings and realizing that one has to face up to the consequences and develop things from there rather than wishing the outcome was different.