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3 years ago

In preparation of aspirin why is water added

1 answer:

8 0

The formation of aspirin will proceed faster if acetic anhydride is used in place of acetic acid.

However, acetic anhydride will hydrolyze in the presence of water to form acetic acid, slowing down the reaction.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Cell notation will list each half-reaction:

likoan [24]

Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

6 0

3 years ago

A sample of pure NO2 is heated to 335 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+

vodka [1.7K]

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t 1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= $\frac{w}{W}$ = $\frac{w}{V}$ = density

P = $\frac{density RT}{M}$ density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = $\frac{0.525 x 0.0821 x 608.15 }{M}$

M = 34.61

to calculate the Kc

Kc= $\frac{ [NO] [O2]}{NO2}$

$\frac{1-2x}{1+x}$ x M NO2 + $\frac{2x}{1+x}$ M NO+ $\frac{x}{1+x}$ M O2

Putting the values as molecular weight of NO2, NO,O2

$\frac{46(1-2x) +30(2x)+32x}{1+x}$

34.61= $\frac{46}{1+x}$

x= 0.33

Kc= $\frac{4x^2)x}{1-2x^2}$

putting the values in the above equation

Kc = 0.00662

5 0

3 years ago

Why potassium by sulphite or KHSO4 donot preserve meat

Debora [2.8K]

Answer:

Postassium bisulphate - KHSO4 - can be used as an food preservative to protect the food from micro-organisms.

4 0

3 years ago

How many grams of NaOH are produced from 1.20 x 10^2 of Na2O

oksian1 [2.3K]

Write an balance the equation

Na2O + H2O -> 2 NaOH

Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.

Na = 23
O=16
H=1

Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40

For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g

120 g Na2O x 80g NaOH / 62g Na2O=

154.8 g NaOH

5 0

3 years ago