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How many molecules (not moles) of \(\rm NH_3\) are produced from 8.69×10−4 \({\rm g}\) of \(\rm H_2\)?
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- Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) 1. What is the % m/m of the solution? 2. What is the mole fraction of the solvent? 3. What is the molarity of the solution? Based from the calculated molarity of the solution in #3, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? and; 4. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) A. What is the %m/m of the solution? B. What is the molarity of the solution? C. What is the mole fraction of the solvent?Write the empirical formula for the hydrated KAl(SO4)2, based on moles of anhydrous KSI(SO4)2= 0.046 mol molar mass of H2O= 18g/ mol moles of H2O= 0.0444mol ratio pf moles H20 to moles of anhydrous KAI(SO4)2= 9.65/1 . Show all work including units. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(SO4)2•4H2O.
- For the following reactions, determine the value of x (in mol)Using the provided data table, calculate ΔrxnH for the following reactions:a) 2O3(g) ⇌ 3O2(g)b) H2S(g) + 3/2 O2(g) ⇌ H2O(ℓ) + SO2(g) Compound ΔfH (kJ mol–1)H2O(ℓ) –285.83H2S (g) –20.6O2 (g) 0O3 (g) 142.7SO2 (g) –296.81Given (8.79x10^6) moles of glucose, C6H12O6, calculate the mole of carbon dioxide, CO2, formed upon complete oxidation of the glucose. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Express your answer in three significant digits as moles. Note: Your answer is assumed to be reduced to the highest power possible.
- {Volume of NaOH dispensed (mL) 41.87mL, 41.95mL, 41.75mL} {Vinegar volume (mL) 5.00mL, 5.00mL, 5.00mL} {NaOH molarity(M) .103M} Average volume of NaOH in liters (L) Average moles of NaOH (mol NaOH) Average moles of acetic acid (mol CH3COOH) Average molarity of acetic acid (M) Average mass of acetic acid (g CH3COOH)…Show that the following reaction obeys the law of conservation of mass assuming there is 0.300g of Al(OH)3. Al(OH)3 + 3HCl -> AlCl3 + 3H2O2N2(g) + 3O2(g) ® 2N2O3(g) DH°rxn = +86.6 kJ 1. When 12.0 moles of N2 react, the reaction heat is ____________ kJ 2. To make +552 kJ of reaction heat, _________ moles of O2 must react. 3. To make +24 kJ of reaction heat, a count of ________________ N2O3 molecules must form.