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Tasya [4]
3 years ago
14

Calculate the empirical formula for DCBN, given that the percent composition is 48.9%C, 1.76%H, 41.2%Cl, and 8.15%N.

Chemistry
1 answer:
anygoal [31]3 years ago
8 0

Answer:

C₇H₃Cl₂N

Explanation:

To determine the empirical formula of DCBN, we need to follow a series of steps.

Step 1: Divide the percent composition of each element by its molar mass

C: 48.9/12.01 = 4.07

H: 1.76/1.01 = 1.76

Cl: 41.2/35.45 = 1.16

N: 8.15/14.01 = 0.582

Step 2: Divide all the numbers by the smallest one

C: 4.07/0.582 ≈ 7

H: 1.76/0.582 ≈ 3

Cl: 1.16/0.582 ≈ 2

N: 0.582/0.582 = 1

The empirical formula of DCBN is C₇H₃Cl₂N

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What is the equilibrium constant of pure water at 25C?
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In pure water, at 25C, the [H3O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of Kw at 25C is therefore 1.0 x 10-14. Although Kw is defined in terms of the dissociation of water, this equilibrium constant expression is equally valid for solutions of acids and bases dissolved in water.

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A mixture of benzoic acid and naphthalene was dissolved in tert-butyl methyl ether and the resulting solution was extracted with
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The sodium benzoate  is completely soluble in the aqueous layer being a salt, and  reacts with hydrochloric acid ( again an acid base reaction ) which precipitates the benzoic acid  since the it is insoluble in water hence  separating it.

5 0
4 years ago
What is the atomic number of the element located in group 16
laila [671]

Answer:

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7 0
3 years ago
Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate
katen-ka-za [31]

The given chemical equation is:

Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)

On balancing the equation we get,

Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

ΔH_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] -   [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of Al_{2}(SO_{4})_{3}reacts will be=

15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol

4 0
4 years ago
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