{"id":256785,"date":"2025-07-16T13:27:06","date_gmt":"2025-07-16T13:27:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256785"},"modified":"2025-07-16T13:27:08","modified_gmt":"2025-07-16T13:27:08","slug":"please-answer-the-following-questions-clearly-and-concisely-using-english-sentences-math-equations-pictures-and-diagrams-as-necessary","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/please-answer-the-following-questions-clearly-and-concisely-using-english-sentences-math-equations-pictures-and-diagrams-as-necessary\/","title":{"rendered":"Please answer the following questions clearly and concisely using English sentences, math equations, pictures, and diagrams as necessary."},"content":{"rendered":"\n

CHM 109 Fall 2020 General Chemistry Laboratory | Pre-Lab Assignment 11: (10 points) Name: Please answer the following questions clearly and concisely using English sentences, math equations, pictures, and diagrams as necessary. The objective is to convey your thoughts as clearly, yet briefly as possible. You may either print this out and complete it on this sheet of paper or complete the assignment on a separate sheet of paper. (3 points) If a solution of hydrogen peroxide (H2O2) is 2.8% (w\/v), that means that for every 100 g of solution, there are 2.8 g of H2O2. What is the molarity of the solution? (Assume the solution has the same density as pure water.) (3 points) If 2.5 mL of the 2.8% H2O2 solution from the previous question reacts according to the following chemical equation, how many moles of water (H2O) and oxygen (O2) will be formed? 2H2O2 -> 2H2O + O2 (3 points) If enough H2O2 decomposes to produce 0.412 moles of O2, and it is 21.0 \u00c2\u00b0C in the lab with an atmospheric pressure of 0.985 atm, what volume will the O2 gas occupy in liters? (1 point) Would you like a free point? Please answer “yes” or “no”.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here are the complete and correct answers to each part of the Pre-Lab Assignment 11:<\/p>\n\n\n\n

\n\n\n\n

(1) Molarity of 2.8% (w\/v) H\u2082O\u2082 solution:<\/strong><\/p>\n\n\n\n

A 2.8% weight\/volume (w\/v) solution means 2.8 grams of H\u2082O\u2082 are present in every 100 mL of solution. To calculate the molarity, use the formula:Molarity=moles of soluteliters of solution\\text{Molarity} = \\frac{\\text{moles of solute}}{\\text{liters of solution}}Molarity=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n

Step 1: Molar mass of H\u2082O\u2082 = 2(1.008) + 2(16.00) = 34.016 g\/mol
Step 2: Convert grams to moles:Moles of H\u2082O\u2082=2.8 g34.016 g\/mol\u22480.0823 mol\\text{Moles of H\u2082O\u2082} = \\frac{2.8\\text{ g}}{34.016\\text{ g\/mol}} \\approx 0.0823\\text{ mol}Moles of H\u2082O\u2082=34.016 g\/mol2.8 g\u200b\u22480.0823 mol<\/p>\n\n\n\n

Step 3: Convert 100 mL to liters = 0.100 L
Step 4: Calculate molarity:Molarity=0.08230.100=0.823 M\\text{Molarity} = \\frac{0.0823}{0.100} = 0.823\\text{ M}Molarity=0.1000.0823\u200b=0.823 M<\/p>\n\n\n\n

Answer: 0.823 M<\/strong><\/p>\n\n\n\n

\n\n\n\n

(2) Moles of water and oxygen from 2.5 mL of solution:<\/strong><\/p>\n\n\n\n

Step 1: Use the molarity from part 1
Step 2: Volume in liters: 2.5 mL = 0.0025 L
Step 3: Moles of H\u2082O\u2082:Moles=0.823 M\u00d70.0025 L=0.00206 mol H\u2082O\u2082\\text{Moles} = 0.823\\text{ M} \\times 0.0025\\text{ L} = 0.00206\\text{ mol H\u2082O\u2082}Moles=0.823 M\u00d70.0025 L=0.00206 mol H\u2082O\u2082<\/p>\n\n\n\n

Step 4: From the balanced equation
2 H\u2082O\u2082 \u2192 2 H\u2082O + O\u2082<\/p>\n\n\n\n

This means for every 2 moles of H\u2082O\u2082, you get 2 moles of H\u2082O and 1 mole of O\u2082.Moles of H\u2082O=0.00206 mol\u00d722=0.00206 mol\\text{Moles of H\u2082O} = 0.00206 \\text{ mol} \\times \\frac{2}{2} = 0.00206\\text{ mol}Moles of H\u2082O=0.00206 mol\u00d722\u200b=0.00206 molMoles of O\u2082=0.00206 mol\u00d712=0.00103 mol\\text{Moles of O\u2082} = 0.00206 \\text{ mol} \\times \\frac{1}{2} = 0.00103\\text{ mol}Moles of O\u2082=0.00206 mol\u00d721\u200b=0.00103 mol<\/p>\n\n\n\n

Answer: 0.00206 mol H\u2082O, 0.00103 mol O\u2082<\/strong><\/p>\n\n\n\n

\n\n\n\n

(3) Volume of O\u2082 gas produced (from 0.412 mol):<\/strong><\/p>\n\n\n\n

Use the Ideal Gas Law:PV=nRTPV = nRTPV=nRT<\/p>\n\n\n\n

Where:
P = 0.985 atm
n = 0.412 mol
R = 0.0821 L\u00b7atm\/mol\u00b7K
T = 21.0\u00b0C = 294.15 KV=nRTP=(0.412)(0.0821)(294.15)0.985V = \\frac{nRT}{P} = \\frac{(0.412)(0.0821)(294.15)}{0.985}V=PnRT\u200b=0.985(0.412)(0.0821)(294.15)\u200bV\u22489.9580.985\u224810.11 LV \\approx \\frac{9.958}{0.985} \\approx 10.11 \\text{ L}V\u22480.9859.958\u200b\u224810.11 L<\/p>\n\n\n\n

Answer: 10.11 L<\/strong><\/p>\n\n\n\n

\n\n\n\n

(4) Free point?<\/strong><\/p>\n\n\n\n

Answer: Yes<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation <\/strong><\/p>\n\n\n\n

In question 1, we calculated molarity using the definition of percent weight per volume. Since the density of the solution is assumed to be the same as water, 100 mL of solution is treated as 100 g. Converting grams of H\u2082O\u2082 to moles using its molar mass and dividing by the volume in liters gives the molarity.<\/p>\n\n\n\n

In question 2, we scaled down the amount from 100 mL to 2.5 mL and used the molarity to find the number of moles of H\u2082O\u2082 present. With the balanced chemical equation, we applied mole ratios to determine how many moles of water and oxygen would be produced. The reaction shows that 2 moles of H\u2082O\u2082 yield 2 moles of water and 1 mole of oxygen, so the ratios are straightforward.<\/p>\n\n\n\n

In question 3, we used the Ideal Gas Law to calculate the volume that 0.412 moles of oxygen gas would occupy under given lab conditions. The temperature had to be converted to Kelvin, and the known gas constant was used. The result shows that a significant volume of oxygen gas is produced even from less than half a mole.<\/p>\n\n\n\n

Lastly, in question 4, answering \u201cyes\u201d secures a free point as instructed.<\/p>\n\n\n\n

Each part builds on chemical understanding, unit conversions, mole relationships, and ideal gas behavior, which are key concepts in general chemistry.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

CHM 109 Fall 2020 General Chemistry Laboratory | Pre-Lab Assignment 11: (10 points) Name: Please answer the following questions clearly and concisely using English sentences, math equations, pictures, and diagrams as necessary. The objective is to convey your thoughts as clearly, yet briefly as possible. You may either print this out and complete it on […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-256785","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256785","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=256785"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256785\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256785"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256785"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256785"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}