please hep with clear steps ( SO I can learn how to do this)
1. If 5.00 g of pure Na2CO3is treated with HCl solution, how many grams of NaCl will be produced if the water is subsequently evaporated from the salt? Molar Mass ofNa 2CO3= 105.99 g/mol.
2. If 5.00 g of pure NaHCO3 is treated similarly, how many grams of NaCl will result?Molar Mass of NaHCO3= 84.01g/mol.
3. What volume of 3.00 M HCl would be required to react with 5.00 g of a mixture that is 50.0% by mass sodium carbonate and 50.0% sodium hydrogen carbonate?
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4. Sodium carbonate (Na:CO) is available in a very pure form and can be used to accurately determine the concentration of an acid solution by titration. This process is called standardization. (15 points total) a. Write a balanced equation for the neutralization of the strong acid HCl by sodium carbonate. The products are a salt, carbon dioxide and water b. You perform a titration experiment and discover it takes 0.265 g of Na:CO, (molar mass 105.99 g/mol) to neutralize 28.3...
How many grams of the primary standard sodium carbonate, Na2CO3 (molar mass=105.99g/mol) is needed to react with 19.4 ml of 0.15 M HCL?
How do I solve this problem step by step? 1.) Calculate the mass in grams) of CO2 produced by the reaction of 0.130 mol of HCl with excess NaHCO3. HCl(aq) + NaHCO3(aq) NaCl (aq) + H20 (1) + CO2(g)
Question 1 According to the following reaction, what mass of silver nitrate would be required to react with 0.500 grams of potassium chloride? AgNO3 (aq) + KCl (aq) --> AgCl (s) + KNO3 (aq) options 2.68 g 0.500 g 85.0 g 170 g 1.14 g Question 2 Consider the reaction: Na2CO3 (aq)+ 2 HCl (aq) --> 2 NaCl (aq) + CO2 (g) + H2O (l) If 43.41 g of sodium carbonate react completely, how many grams of HCl will be...
A 0.6739 g sample of a pure carbonate, X,CO,(s), was dissolved in 50.0 mL of 0.1850 M HCl(aq). The excess HCl(aq) was back titrated with 24.70 mL of 0.0980 M NaOH(aq). How many moles of HCl react with the carbonate? moles of HCI = mol What is the identity of the cation, X? cation: A standardized solution that is 0.0100 M in Na+ is necessary for a Hame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate...
#4.. please show work and dont skip steps. trying to learn how to do this for the actual biochemistry lab PART B. PRACTICE CALCULATIONS Convert units (show your work!): a. 0.01 ml b. 0.05 M = MM c. 0.2 M = mM d. 10 mM - M 2. Molar mass of Na olar mass of NaOH is 40 /mol. How many grams of NaOH do you have to take to prepare a. 11 of 10 M solution of NaOH? b....
A sample of hydrate was heated thoroughly. How many waters of hydration are there if the anhydrous salt had a dry mass of 12 grams and a molar mass of 185 g/mol? The mass of water evaporated from the hydrate was 5.0 g.
3) Convert 0.200 g of sodium carbonate and 0.200 g of sodium bicarbonate to moles. Molar Mass of sodium carbonate 106.0 g/mol Molar Mass of sodium bicarbonate 84,0 g/mol 4) An unknown compound, "X", is known to be either SrCl2 or SrBr2. A 12.0 g sample of this compound yields 4.25 g of strontium. Identify the sample. Show the work. 5) How many milliliters of hydrochloric acid are required to react with 0.20 g of sodium carbonate? You must show...
Table Salt from Baking Soda Name: Date: Partner: DATA: Mass of beaker 95.85. Mass of beaker and sample (NaHCO3) 17.45. Mass of sample (NaHCO3) Mass of beaker and residue (NaCl) 1" Heating... :26 99, Mass of beaker and residue (NaCl) 2 Heating.. 26.85, Mass of beaker and residue (NaCl) 3* Heating... 26.959 Mass of beaker and residue (NaCl) best g Mass of residue (NaCl) 9 Volume of conc. HCI added ml CALCULATIONS: g/mol Molar mass of NaHCO3 Number of moles...
1. 5.00 g of butane are combusted in a sealed container containing 2.00 g of oxygen gas. The balanced reaction for the combustion of butane is shown below. How many grams of the excess reagent are left over? (This question first requires you to determine the limiting reagent, and then calculate how much will be leftover). Pease answer in g. Molar Mass Butane: 58.12 g/mol Molar Mass Oxygen Gas: 32.0 g/mol 2CH3CH2CH2CH3 + 13O2 --> 8CO2 + 10H2O 2. If...