Try Another Version Submit Answer 3 item attempts remaining A chemist wants to prepare 0.740 M HCI, Commercial hydrochloric acid is 12.4 M. How many milliliters of the commercial acid does the chemist require to make up 1.40 L of the dilute acid? [References]

I Review | Constants | Periodic Table You may want to reference (Pages 298 - 306) Section 9.4 while completing this problem. Part A Calculate the molarity of the following solutions. Calculate the molarity of 0.550 mol of glucose in 0.350 L of a glucose solution. Express your answer with the appropriate units. HA ? molarity = Value Units Submit Request Answer Part B Calculate the molarity of 89.5 g of HCl in 1.50 L of an HCl solution. Express your answer with the appropriate units. HA ? molarity = Value Units

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Calculate Average molarity (M) of Acetic acid in vinegar, using the following data. Calculate the mass percent of acetic acid in vinegar, using your average Acetic acid concentration (M) from above result. Table 2. Data for titration of 20.00 mL of vinegar with approximately 1.2 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH.

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II Calculating molarity using solute moles A chemist prepares a solution of iron(II) bromide (FeBr,) by measuring out 152. umol of iron(II) bromide into a 200. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's iron(II) bromide solution. Round your answer to 3 significant digits. mol Explanation Check O 2020 McGraw-Hill Education. All Rights Reserved. Terms of Use | Privacy l Accessi Type here to search 番 き

Data Table Final buret reading Initial buret reading Volume used Sample 1 Vinegar Base Vinegar Sample 2 Sample 3 (if needed) Base Vinegar Base JOML Som 0,05 110mL/802 10ML 884 JUML 0.00 10μL OML Tome 0.00 10mL 80.2 mL 10ML 88mL come some *If a pipet is used to measure the volume of vinegar, record only in the space for volume used. Molarity of base (NaOH) 1685 M Vinegar number N/A CALCULATIONS: In the spaces below, show calculation setups for Sample 1 only. Show answers for both samples in the boxes. 1. Moles of base (NaOH) 2. Moles of acid (HC2H3O2) used to neutralize (react with) the above number of moles of base 3. Molarity of acetic acid in the vinegar Sample 1 Sample 2 Sample 3 (if needed) 4. Average molarity of acetic acid in the vinegar 5. Grams of acetic acid per liter (from average molarity) 6. Mass percent acetic acid in vinegar sample (density of vinegar = 1.005 g/mL) - 252 -