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Jlenok [28]
3 years ago
7

Which of these is not an example of a possible additive used in ink?

Chemistry
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

greases

Explanation:

Ink is manufactured in two stages: first varnish (a mixture of solvent, resins and additives) is made and then pigments are mixed into it. Varnish is a clear liquid that solidifies as a thin film. It binds the pigment to the printed surface, provides the printability of the ink and wets the pigment particles. Ink made with color usually includes the following ingredients: petroleum distillate solvent, linseed oil, some form of organic pigments, and soybean oil. Inorganic pigments are not commonly used.

Additives play a critical role in the formulation and production of inks, helping to impart the necessary characteristics the ink needs, whether it is to run more smoothly on press or to protect the finished product.

However, grease is not one of the likely additives to ink hence it is not essentially used in ink production.

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3 years ago
The solubility of CaCl2 in water is 25g per 50 mL of H2O. How many grams of CaCl2 will dissolved to form a saturated solution in
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Answer: option B. 5g

Explanation:

50mL of H2O dissolves 25g of CaCl2.

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4 years ago
When 412.5g of calcium carbonate react with 521.9g of aluminum fluoride,how many grams of each product can be produced
aliya0001 [1]
<h3><u>Answer;</u></h3>

321.8 g CaF2

321.5 g Al2(CO3)3

<h3><u>Explanation;</u></h3>

The equation for the reaction is;  

3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3  

Number of moles of CaCO3 will be;

=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)

= 4.12139 mol CaCO3  

Number of moles of AlF3 will be;

= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)

= 6.21482 mol AlF3

But;

4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.

Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.  

Therefore;

Mass of CaF2 will be;

(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2  

Mass of Al2(CO3)3 on the other hand will be;

(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3

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