Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The meat from one hazelnut has a mass of \(0.985 \mathrm{~g}\). (a) What is the mass of a millionth of a mole \(\left(10^{-6}\right)\) of hazelnut meats? (A millionth of a mole is also called a micromole.) (b) How many moles are in a pound of hazelnut meats?

Short Answer

Expert verified
Answer: The mass of a micromole of hazelnut meats is \(970 \times 10^{-9}\, \text{g}\), and there are approximately \(460.9\, \text{mol}\) of hazelnut meats in a pound.

Step by step solution

01

(Part a: Find the mass of a micromole of hazelnut meats)

We follow these steps to find the mass of a micromole of hazelnut meats: 1. Determine the moles of one hazelnut meat (\(n_\text{meat}\)), given that one hazelnut has a mass of \(0.985 \mathrm{~g}\). 2. Divide the number of moles of one hazelnut meat (\(n_\text{meat}\)) by a million to get the moles for a micromole. 3. Multiply the moles of a micromole by the mass of one hazelnut meat to determine the mass of a micromole of hazelnut meats. First, we calculate the moles of one hazelnut meat (\(n_\text{meat}\)) by dividing the mass of one hazelnut meat by its molar mass: $$n_\text{meat} = \frac{m_\text{meat}}{M_\text{meat}}$$ We need the molar mass of hazelnut meat, but it's not provided. In this case, we assume that the mass of a hazelnut corresponds to one unit, so the molar mass is 1 g/mol. $$n_\text{meat} = \frac{0.985 \mathrm{~g}}{1} = 0.985\, \text{mol}$$ Now, we find the moles of a micromole by dividing the moles of one hazelnut meat by a million: $$n_\text{micromole} = \frac{n_\text{meat}}{10^6} = \frac{0.985\, \text{mol}}{10^6} = 9.85\times10^{-7}\, \text{mol}$$ Finally, we determine the mass of a micromole of hazelnut meats: $$m_\text{micromole} = n_\text{micromole} \cdot m_\text{meat} = 9.85\times10^{-7}\, \text{mol} \cdot 0.985 \mathrm{~g} = 970\times10^{-9}\, \text{g}$$ Thus, the mass of a micromole of hazelnut meats is \(970 \times 10^{-9}\, \text{g}\).
02

(Part b: Calculate the number of moles in a pound of hazelnut meats)

First, we need to convert the mass of a pound of hazelnut meats to grams. There are 454 grams in a pound. So, the mass of a pound of hazelnut meats is: $$m_\text{pound} = 1\, \text{lb} \cdot \frac{454\, \text{g}}{1\,\text{lb}} = 454\, \text{g}$$ Now, we calculate the number of moles in a pound of hazelnut meats by dividing the mass of a pound of hazelnut meats by the mass of one hazelnut meat: $$n_\text{pound} = \frac{m_\text{pound}}{m_\text{meat}} = \frac{454\, \text{g}}{0.985 \mathrm{~g}} \approx 460.9\, \text{mol}$$ Thus, there are approximately \(460.9\, \text{mol}\) of hazelnut meats in a pound.

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

How many electrons are in (a) an ion of \(\mathrm{Sc}^{3+}\) ? (b) a mol of \(\mathrm{Sc}^{3+}\) ? (c) a gram of \(\mathrm{Sc}^{3+}\) ?

When iron and steam react at high temperatures, the following reaction takes place. $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$ How much iron must react with excess steam to form \(897 \mathrm{~g}\) of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) if the reaction yield is \(69 \%\) ?

Balance the following equations: (a) \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) (b) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the following equations: (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{XeF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Xe}(g)+\mathrm{O}_{2}(g)+\mathrm{HF}(g)\) (c) \(\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{HCl}(g)\)

When copper(II) oxide is heated in hydrogen gas, the following reaction takes place. $$ \mathrm{CuO}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ A copper rod coated with copper(II) oxide has a mass of \(38.72 \mathrm{~g}\). The rod is heated and made to react with \(5.67 \mathrm{~L}\) of hydrogen gas, whose density at the conditions of the experiment is \(0.0519 \mathrm{~g} / \mathrm{L}\). (a) How many grams of \(\mathrm{CuO}\) were converted to Cu? (b) What is the mass of the copper after all the hydrogen is consumed? (Assume that \(\mathrm{CuO}\) is converted only to Cu.)

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