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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Answer:
Step-by-step explanation:
Alright, lets get started.
Suppose they take t minutes to meet each other.
Distance covered by first friend in t minutes, = 0.2 *t=0.2∗t
Distance covered by second friend in t minutes , =0.15 *t=0.15∗t
Total distance is given as 7, so
0.2 t + 0.15 t = 70.2t+0.15t=7
0.35 t = 70.35t=7
t = 20t=20
means after 20minutes they will meet.
SO. the average speed is 10m: Answer
Below are the steps to get the answers:
<span>1.) write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
2.) You are given everything needed to calculate
q=heat transfer=2.2*10^2, H3PO4 moles= 1.5*10^-3, NaOH moles=5.0*10^-3
3.) equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
4.) Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj, </span>
Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)
That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.
