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CaHeK987 [17]
3 years ago
8

A solution of pyridinium bromide has a pH of 3.00. What is the concentration of the pyridinium cation at equilibrium, in units o

f molarity? Express molarity to two decimal places.
Chemistry
1 answer:
Artemon [7]3 years ago
4 0

Answer:

The concentration of the pyridinium cation at equilibrium is 1.00×10⁻³ M

Explanation:

In water we have

C₅H₅NHBr + H₂O ⇒ C₅H₅NH+ + Br−

Pyridinium Bromide (C₅H₅NHBr) Dissociates Completely  Into C₅H₅NH+ And Br− as such it is a strong Electrolyte.

Therefore the number of moles of positive ion produced per mole of C₅H₅NHBr  is one

pH = - log [H₃O⁺] Therefore 10^-pH = [H₃O⁺] = concentration of C₅H₅NHBr

= 10⁻³ = 0.001M = concentration of C₅H₅NHBr

The concentration of C₅H₅NHBr is = 1.00×10⁻³ M to two places of decimal

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To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o
Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

\pi  r^{2}h ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

r=\frac{d}{2}

=\frac{3.36x10^{2}}{2}

= 168 m

Volume =

=\frac{22\times 168^{2}\times 21.80}{7}

=1.93\times 10^{6} m^{3}

2.Convert ppm to g/m3 and Solve for mass of F-

1ppm = 1g/m^{3}

0.9ppm = 0.9g/m^{3}

Because both ppm and g/m3 are same quantity .

g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}

0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}

mass\ of\ F- =1.740g

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

2Na+F_{2}\rightarrow 2NaF

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g  of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

\frac {84}{38}\times 1.74

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

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3 years ago
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Answer:

in the simple cubic unit cell, the centers of ____Eight________ identical particles define the ____corners________ of a cube. The particles do touch along the cube's _____edges _______ but do not touch along the cube's ___ diagonally_________ or through the center. There is/are __eight__________ particle per unit cell and the coordination number is _____six_______ .

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A nitrde ion has 7protons,8nutrons and 10 electrons. what is the overall change
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Answer:

the overall charge on the nitride anion is  

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N power 3 − → the nitride anion

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