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3 years ago

What are the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry

Chemistry

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer:

Organic

Plate making

Ink mists

Gas, fumes and dust

Explanation:

the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry are Organic ,Plate making ,Ink mists, Gas, fumes and dust.

Organic refer to those chemicals gotten from living matter and are natural.

Plate making refers to how plates are make and it carry image in the printing process.

Ink mist gotten from ink is use in graphic designs for painting and color indication.

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Answer:

The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.

Explanation:

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$HA + H_{2}O$ ⇄ $A^{-} + H_{3}O^{+}$

This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,

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Explanation:

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Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

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