How many moles in 200 grams of Br

(1) Write the net ionic equation for the reaction that occurs when equal volumes of 0.233 M aqueous acetylsalicylic acid (aspiri

AveGali [126]

Complete Question

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Answer:

1) The ionic equation is

$HC_9H_7O_4 + (C_2 H_5)_3 N -----> C_9H_7O_4^- + (C_2 H_5)_3 NH^-$

2

At equilibrium The aniline will be favored

3

The pH of the solution is $pH = 7.12$

Explanation:

From the question we are told that

The concentration of aspirin is $C_A = 0.233M$

The acid dissociation constant for aspirin is $K_a = 3.0*10^{-4}$

The base dissociation constant for aniline is $K_b = 7.4 *10^{-10}$

The molecular formula for aspirin is $HC_9H_7O_4$

The molecular formula for aniline is $(C_2 H_5)_3 N$

So the net ionic reaction is

$HC_9H_7O_4 + (C_2 H_5)_3 N -----> C_9H_7O_4^- + (C_2 H_5)_3 NH^-$

Generally pKa is mathematically evaluated as

$pKa = -log(3.0*10^{-4})$

$= 3.52$

Generally pKb is mathematically represented as

$pKa = -log(7.4*10^{-10 })$

$= 9.13$

Generally

pKa + pKb = 14

So for the triethylammonium salt produced the pKa is

$pK_a__{s}} = 14 - 3.28$

$= 10.72$

As a result of the pKa of aspirin being lower that that of triethylammonium salt at equilibrium it implies that aniline would be favored

The pH of the solution is mathematically represented as

Since they are of equal mole

$pH = \frac{1}{2} (pKa + pKb + pKw)$

Where pKw is the pKw of water which has a value of 14

$pH = 0.5(14 + 3.52 + 3.28)$

$pH = 7.12$

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If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

$\tt \dfrac{8.5}{78.11}=0.109$

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

$\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol$

Heat transferred to water :

Q=m.c.ΔT

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If a ball is on the top of a hill, it has the most potential energy on the very top of the hill. The kinetic energy is also 0 at this point.

If the ball rolls down the hill, potential energy decreases while kinetic increases.

simple example

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