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Which product is used on the natural nail prior to product application to assist in adhesion and serves to chemically bond the e

ira [324]

The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

<h3>What is a nail primer?</h3>

A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.

The nail primers are also very useful for improving the cleaning efficiency of the product before its application.

Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.

In conclusion, the chemical formulation employed on the natural nail that is capable of enhancing and also assisting adhesion is called the nail primer.

Learn more about nail esthetic products here:

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8 0

2 years ago

Why does atomic radii decrease as you move from the left to right across the period

artcher [175]

Answer:

This is because, within a period or family of elements, all electrons are added to the same shell.

5 0

2 years ago

Read 2 more answers

What are the three types of protist and how are they similar and how are they different (this is science but there is no tab for

allochka39001 [22]

Answer:

They are eukaryotic, which means they have a nucleus. Most have mitochondria

3 0

3 years ago

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

riadik2000 [5.3K]

Answer:

A) SiO2 is the limiting reactant

B) Theoretical yield= 72333.3g

C) % yield =91.5%

Explanation:

SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)

n(SiO2)= 155000/60 = 2583.33 mols

n(C)= 79000/12= 3291.66 mols

a)SiO2 is the limiting reactant

According to the balanced reaction equation,

60g of SiO2 produced 28g of SiO2

155000g of SiO2 will produce 155000×28/60= 72333.3g

Therefore theoretical yield of Si= 72333.3g

% yield= 66200/72333.3×100/1 =91.5%

5 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago