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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1.94 moles
I did 35/18.02 because 18.02 is the molar mass of water
Answer:
178.98 sq. feet
Explanation:
The path and the garden has been shown in the figure below. The green area is the garden and the area in brown is the path.
It has been given that,
Radius of garden = 8 feet
So, the area of garden = 3.14 × 8 × 8 = 200.96 sq. feet
The total radius of the land including garden and path = 8 + 3 = 11 feet
So, the total are of land including garden and path = 3.14 × 11 × 11 = 379.94 sq. feet
So, the area of path = Total area of the land - area of garden
Area of path = 379.94 - 200.96 = 178.98 sq. feet
Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)
moles KOH = 4 x 2 / 1
= 8 moles of KOH
molar mass KOH = 56 g/mol
mass of KOH = n x mm
mass of KOH = 8 x 56
= 448 g of KOH
hope this helps!
Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
