The volume of a red blood cell is about $90.0 \times 10^{-12} \mathrm{cm}^{3} .$ Assuming that red blood cells are spherical, what is the diameter of a red blood cell in millimeters?

Short Answer

Expert verified
The diameter of a red blood cell is approximately 0.0554 mm.

Step by step solution

01

Re-arrange the formula for sphere volume to solve for r

Firstly, we need to make 'r' the subject of the formula, which gives us \(r = (\frac{3v}{4\pi})^{\frac{1}{3}}\).
02

Substitute the volume of the red blood cell into the formula

Given that the volume of the red blood cell is \(90.0 \times 10^{-12} cm^{3}\), replacing 'v' with this value in the formula gives \(r = (\frac{3(90.0 \times 10^{-12})}{4\pi})^{\frac{1}{3}}\).
03

Calculate the radius r

Evaluating the expression for 'r' gives us a result in centimeters (approximately 0.00277 cm).
04

Convert radius to diameter in millimeters

Since diameter is twice the radius, the diameter in centimeters is approximately 0.00554 cm. Conversion of this to millimeters (by multiplying with 10 because 1 cm = 10 mm) gives an approximate diameter of 0.0554 mm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a user's manual accompanying an American-made automobile, a typical gauge pressure for optimal performance of automobile tires is $32 \mathrm{lb} / \mathrm{in} .^{2} .$ What is this pressure in grams per square centimeter and kilograms per square meter?

A vinegar sample is found to have a density of \(1.006 \mathrm{g} / \mathrm{mL}\) and to contain \(5.4 \%\) acetic acid by mass. How many grams of acetic acid are present in \(1.00 \mathrm{L}\) of this vinegar?

In the third century \(\mathrm{BC}\), the Greek mathematician Archimedes is said to have discovered an important principle that is useful in density determinations. The story told is that King Hiero of Syracuse (in Sicily) asked Archimedes to verify that an ornate crown made for him by a goldsmith consisted of pure gold and not a gold-silver alloy. Archimedes had to do this, of course, without damaging the crown in any way. Describe how Archimedes did this, or if you don't know the rest of the story, rediscover Archimedes's principle and explain how it can be used to settle the question.

Indicate whether each sample of matter listed is a substance or a mixture; if it is a mixture, indicate whether it is homogeneous or heterogeneous. (a) a wooden beam (b) red ink (c) distilled water (d) freshly squeezed orange juice

A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at \(25^{\circ} \mathrm{C}\), and the density of water at \(25^{\circ} \mathrm{C}\) is \(0.9970 \mathrm{g} / \mathrm{mL}\) ). At the left, grains of the mineral calcite float on the surface of the liquid bromoform \((d=2.890 \mathrm{g} / \mathrm{mL})\) At the right, the grains sink to the bottom of liquid chloroform \((d=1.444 \mathrm{g} / \mathrm{mL}) .\) By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined (Exercise 62).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free