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2 years ago

A molecule has an empirical formula of ch, and its molar mass is known to be 26 g/mol. What is its molecular formula?.

1 answer:

7 0

A molecule has an empirical formula of ch, and its molar mass is known to be 26 g/mol and the molecular formula is C₂H₂ ethyne

Molecular formula of compound is (CH)n and the given molar mass is 26g/mol

Molar mass of (CH)n, C=12=n(12+1)=13n

So 13n and n=2

=13×2=26 and given molar mass is also 26g/mol

So here two carbon and two hydrogen so molecular formula is C₂H₂ and name is ethyne

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There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen

alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

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3 years ago

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3 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

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3 years ago