Sodium hydrogen carbonate (NaHCO3), also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid (HCI), which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO₂(9) The CO2 gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 250. mL of a 0.076 M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCI ? Round your answer to 2 significant digits.

using this balanced equation CaCO3(s) + 2 HCl(aq) ----> H2O(l) + CO2(g) + CaCl2(aq) Stomach acid is 0.100 M HCL. An active ingreident found in antacids such as Alka Seltzer is calcium carbonate, CaCO3. Calculate grams of calcium carbonate needed to react with 80 ml of stomach acid.

A) A student completely dissolves 5.16 mol of Ca(OH)2 in water. i) Which equation is the dissociation equation for Ca(OH), in water? (Give the letter choice)| A. Ca(OH) 2 (s) + H 20 ---> CaO(aq) + O 2 (g) + 2 H 2 (g) С. Са(ОН) 2 (s) --> Са 2+ (aq) + 20 2- (aq) + 2H * (aq) В. Cа(ОН) 2 (s) ---> Са 2" (аq) + 2 ОН (аq) D. Ca(ОН) 2 (s) --> CаН 2 (aq) + O 2 (g) ii) How many total moles of ions are released from the 5.16 mol of Ca(OH)2 ? mol of ions B) A sample of FeCl3 is completely dissolved in water. i) Which is the dissociation equation for FeCl3 in water? (Give the letter choice) A. FeCl 3 (s) + H 20 ---> FeCl 2 (aq) + HCI(aq) + OH (aq) C. FeCl (s) ---> Fe 3+ (aд) + Cl 2 (aq) + Cl (aq) 3 В. FeCl 3 (s) + Н20 ---> FEHCI (aq) + ОН " (aq) D. FeCl 3 (s) ---> Fe 3+ (ад) + 3 CI (аq) ii) How many total moles of ions are released if 5.71 x1019 formula units of FeCl3 were completely dissolved? (NOTE: Write answer in sci. notation using "E" or "e" to replace "x 10" & do NOT leave any space between…

A 6.404 gram sample of sodium hydroxide (NaOH) is mixed with 22.50 mL of a 0.642 M HCl solution. Determine if excess acid or base is present after being mixed.  Determine how many moles of excess acid or base is present.  Determine the molarity of excess acid or base.  Finally determine the pH of the solution.   NaOH(s)  +  HCl(aq)  →  H2O(L)  +  NaCl(aq)

The end point in a titration of 50.00 mL of aqueous NaOH was reached by the addition of 35.23 mL of 0.250 M aqueous HCI titrant. Calculate the molarity of the NaOH solution. The titration reaction equation is: NaOH(aq) + HCI (aq) —> NaCl(aq) + H2O(l)

Complete the equation for the dissociation of Fe(ClO4)3(aq). Omit water from the equation because it is understood to be present. Fe(ClO4)3(aq) -->

A technician would like to standardize an unknown Ca(OH)2 solution. It takes 18.53 mL of a 0.174 M solution of HCI to titrate 25.00 mL the unknown Ca(OH)2 solution. Calculate the molarity of the Ca(OH)2 solution. Tip: Don't forget to consider the mole ratio between HCl and Ca(OH)2. 2 HCI (aq) + Ca(OH), (aq) CaCl2 (aq) + 2 H2O (1)

A scientist is investigating the solubility of two polyatomic ions, oxalate (C2O2−4C2⁢O42−) and arsenate (AsO3−4AsO43−), which are not listed in the table of solubility guidelines. The scientist starts with four solutions made of water-soluble salts. Solution A contains sodium arsenate. Solution B contains ammonium oxalate. Soution C contains silver chlorate. Solution D contains aluminum bromide. Solution Solute Color of Solution A Na3AsO4Na3⁢AsO4 colorless B (NH4)2C2O4(NH4)2⁢C2⁢O4 colorless C AgClO3AgClO3 colorless D AlBr3AlBr3 yellow The results of mixing each solution in pairs are shown in the table. Experiment Solutions Mixed Result 1 A + B no precipitate, colorless solution 2 A + C brown precipitate 3 A + D white precipitate 4 B + C white precipitate 5 B + D white precipitate 6 C + D yellow precipitate Identify the formula for each precipitate that forms. precipitate from A+C:   precipitate from A+D:   precipitate…

Consider the reaction: HCI(aq) + NH, (ag) – NH,Cl(aq) Where 20.00 mL of 0.100 M NH, (aq) is added to 10.00 mL of 0.200 M HCl(aq). Order the solution components (excluding water) from highest to lowest concentration. O [OH¯] > [NH3] > [Na*] > [Cl¯] > [H,O*] O (CI ] = [NH†] > [NH,] = [H,O*]> [OH¯] O [H,0*)> [Cl¯] > [NH,] > [NH†]> [OH) O [NH]= [OH"] > [H,O*] = [CI¯] > [NH,] O [H,0*] = [CI¯] > [NH,] > [NH ] = [OH] O [NH,CI] = [NH,] = [HCl] O [CI] > [NH†]> [H,O+] > [NH,] > [OH ] O INH,CI] > [NH3] > [HCl] O None of these

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation:HA(aq)+KOH(aq) --------->KA(aq)+H2O(l)       If 14.90 mL of 0.585 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? ______ M      What is the molar mass of HA? _____  g/mol