I believe the answer to this question is true
Explanation:
From the question it can be infered that initially only parent atoms were present in the rock and no daughter atom. So, initially there was only uranium atom and no lead atom. So, in total there were 500+1500= 2000 uranium atom. And no lead tom was there.
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:
brainly.com/question/14498053
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The question is incomplete. the complete question is:
A chemistry student needs 35.0 g of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of pentane is
. Calculate the volume of pentane the student should pour out.Round your answer to 3 significant digits.
Answer: The volume of pentane the student should pour out is 55.9 ml
Explanation:
Density is defined as the mass contained per unit volume.
To calculate volume of a substance, we use the equation:
![\text{Density of pentane}=\frac{\text{Mass of pentane}}{\text{Volume of pentane}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20pentane%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20pentane%7D%7D%7B%5Ctext%7BVolume%20of%20pentane%7D%7D)
We are given:
Density of pentane = ![0.626g/cm^3](https://tex.z-dn.net/?f=0.626g%2Fcm%5E3)
Mass of pentane = 35.0 g
Putting values in above equation, we get:
![0.626g/cm^3=\frac{35.0g}{\text {Volume of pentane}}\\\\\text{Volume of pentane}=55.9cm^3=55.9ml](https://tex.z-dn.net/?f=0.626g%2Fcm%5E3%3D%5Cfrac%7B35.0g%7D%7B%5Ctext%20%7BVolume%20of%20pentane%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20pentane%7D%3D55.9cm%5E3%3D55.9ml)
(Conversion factor:
)
Hence, the volume of pentane is 55.9 ml