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Anvisha [2.4K]
3 years ago
9

A 0.167-g sample of an unknown compound contains 0.00278 mol of the compound. Elemental analysis of the acid gives the following

percentages by mass: 40.00% C; 6.71% H; 53.29% O.
Chemistry
1 answer:
ratelena [41]3 years ago
8 0

Answer:

Emperical formula of the acid is CH2O

Molecular formula is C2H4O2

Explanation:

C. H. O

% by mass. 40. 6.71 53.29

MW. 12. 1. 16

Mole 40/12. 6.71/1. 53.29/16

3.33. 6.71. 3.33

3.33/3.33 6.71/3.33 3.33/3.33

1. 2. 1

Emperical Formula (EF): CH2O

Molecular Formula (MF)

Molecular Weight (MW) of acid = mole/mass = 0.00278/0.167 = 60g/mol

MF = (EF)n = 60

(CH2O)n = 60

(12+2+16)n = 60

30n = 60

n = 60/30 = 2

Molecular formula = (EF)n = (CH2O)2 = C2H4O2

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What is the [H+] of a solution with a pH of 9.40?
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In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:

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A compound possessing a carboxylic acid produces a yellow solution when dissolved in diethyl ether. This yellow solution is tran
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The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

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k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

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Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

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[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

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Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

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t=147.24\ min

6 0
3 years ago
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