20.26 grams of unknown 1 (with molar mass 203.67 g/mol) was dissolved in 182.7 mL of...

20.26 grams of unknown 1 (with molar mass 203.67 g/mol) was dissolved in 182.7 mL of ethylene glycol in a calorimeter. What change in temperature would the solution experience as a result of this dissolution? Assume the solution has the same specific heat capacity as pure ethylene glycol. The density of ethylene glycol is 1.11 g/cm3. The ∆Hdissolution for unknown 2 is 124.1 kJ/mol unknown 1.

Expert Solution

queen_honey_blossom answered 2 years ago

Determine the molarity when 12.9 grams of ethanol, C2H5OH (Molar mass = 46.08 g/mol) is dissolved...

Determine the molarity when 12.9 grams of ethanol, C2H5OH (Molar mass = 46.08 g/mol) is dissolved in 248.8-mL of solution. Answer to 2 decimal places. Determine the molality when 4.93 moles of ethanol are dissolved in 0.489 kg of water. Answer to 2 decimal places. Determine the molality of a solution when 43.4 grams of ethanol, C2H5OH (Molar mass = 46.08 g/mol) is dissolved in 0.189 kg of water. Answer to 2 decimal places. Determine the molality when 24.3 grams...

Part #1: 30.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) are dissolved in 100....

Part #1: 30.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) are dissolved in 100. grams of water (molar mass 18.02). What is the mole fraction of water in solution? Note: glucose is a non-volatile solute. Part #2: At 25.0 oC, pure water has a vapor pressure of 24.2 torr. What would the vapor pressure of the solution in part one be (in torr)? Part #3: A liquid is a mixture of benzene and toluene. The sample contains 0.225...

How many g of MgCl2 (molar mass = 95.21 g/mol) have to be dissolved to give...

How many g of MgCl2 (molar mass = 95.21 g/mol) have to be dissolved to give 100 g of an aqueous solution that boils at 101.50 degree C? A) 279g B) 27.9g C) 95.2g D) 93.0g E) 9.30g

A 0.1276−g sample of a monoprotic acid (molar mass = 1.10 × 102 g/mol) was dissolved...

A 0.1276−g sample of a monoprotic acid (molar mass = 1.10 × 102 g/mol) was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH. After 10.0 mL of base had been added, the pH was determined to be 4.87. What is the Ka for the acid? Answer in scientific notation.

An unknown compound with a molar mass of 176 g/mol is analyzed and determine to consist...

An unknown compound with a molar mass of 176 g/mol is analyzed and determine to consist of 40.91 mass percent carbon, 4.58 mass percent hydrogen and the remainder oxygen. What are the empirical and molecular formulas for this compound

When 2.65 g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol...

When 2.65 g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of the resulting solution is -0.257 ∘C.

Chemical analysis of an unknown compound gives a molar mass of 334 g/mol and a percentage...

Chemical analysis of an unknown compound gives a molar mass of 334 g/mol and a percentage composition of 75.42% carbon, 6.63% hydrogen, 8.38% nitrogen, and the rest is oxygen. Calculate the empirical and molecular formulas for the unknown compound. 2.) Write a balanced chemical equation for the complete combustion of pentane (C5H12). If 100.00 g of pentane are combusted in excess O2, what mass of CO2 would be released to the atmosphere?

Chem Question - Gases Molar Mass N = 14.01 g/mol Molar Mass H2O = 18.016 g/mol...

Chem Question - Gases Molar Mass N = 14.01 g/mol Molar Mass H2O = 18.016 g/mol Vapor Pressure of Water at 25 C is 23.76 torr Vapor Pressure of Water at 65 C is 187.54 torr (Show all work and calculations, include units in answer. If calculations must be used in several parts, rounding should be made to 6 decimal points to ensure accuracy. Final Answers can be rounded to 3 decimal points.) A syringe filled with air can be...

The reaction below was performed with 0.342 grams of K2CO3 (molar mass = 138.21 g/mol). After...

The reaction below was performed with 0.342 grams of K2CO3 (molar mass = 138.21 g/mol). After the reaction was complete 0.0675 grams of CO2 was recovered. What was the percent yield to 2 significant figures? K2CO3(s) --> K2O(s) + CO2(g) The correct answer is 62 %, but I don't understand how they got there. Please help explain the steps

If the sollubility of sodiym acetate (Molar Mass = 82 g/mol) is 76 grams per 100...

If the sollubility of sodiym acetate (Molar Mass = 82 g/mol) is 76 grams per 100 grams of water, which of the following solutions would be considered supersaturated (holding more than the normal maxiym solution)? a.) 8.5 moles of sodium acetate dissolved in 1 L of water, b.) 1.8 moles of sodiym acetate dissoved in 300 mL of water, c.) 5.5 moles of sodiym acetate dissolved in 500 mL of water, or d.) 1.2 moles of sodium acetate dissoved in...

Question