Who is right, John or Anna? Explain your answer. 3 When 200 g of calcium nitrate, Ca(NO3)2.2H2O is heated at 120ºC the mass decreases by 36 g. (a) Why does the mass decrease? (b) How much calcium nitrate is left after heating? 4 Calculate the molar mass of these compounds: (relative atomic masses: H = 1; N = 14; O = 16; S = 32; Cu = 64; Br = 80; Pb = 207) (a) copper nitrate, Ca(NO3)2 (b) lead bromate, Pb(BrO3)2 (c) ammonium sulfate, (NH4)2SO4 5 The equation for the complete combustion of methane is shown below.
The determined ratio was 1:1.01, Mg:O. Even though the ratio does round to the accepted 1:1 relationship, error likely occurred, and was contributed through loss of product during combustion. While heating, it was difficult to detect smoke. This would lower the calculated mass of oxygen, and its ratio number, while increasing the ratio number for magnesium. Secondly, upon inspection of the contents in the crucible, not all of the contents appeared white, suggesting that possibly not all the magnesium reacted.
Discussion: a. The solubility of KNO3 at 50 ̊ C is approximately 92g/100g. b. The temperature at which the solubility of KNO3 is 80g/100g water is 44 ̊ C. c. There should be 105g of potassium nitrate to dissolve in 100g of water at 60 ̊ C. d. Using the formula of solubility, it came out that the temperature needed was 80 ̊ C to completely dissolve 20g of potassium nitrate in 25g of water. The possible errors may include: * The difference between what one person would determine as a crystal compared to another, the time between when the temperature was recorded and when the
RESULTS The first flask held .305 grams; the second, .454 grams; and the third, .477 grams of unknown gas. According to the ideal gas law, at a pressure of 762.0 mmHg and 16.0ºC, a vessel of 250 ml will contain .0105 moles of gas, while a vessel of 252 ml will contain .0106 moles of gas. Dividing the number of grams of unknown gas contained in each flask by the corresponding number of moles contained in that vessel resulted in a molar mass for each of the flasks. The
What volume of concentrated (18.0 M) sulfuric acid would be required to make each of the following? a. 1.25 L of 6.00 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (6.00 M) x (1.25 L) Vconc=0.417 L b. 575 mL of 0.100 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (0.100 M) x (0.575 L) Vconc=0.00319 L Acid-Base Titrations 5. Calculate the molarity of an HCl solution if 20.0 mL of it requires 33.2 mL of 0.150 M NaOH for neutralization.
Delta T= 18 M= 35 ml Starting Temp: 23 degrees Celsius Final Temp: 41 degrees Celsius q=(35)(4.184)(18) = 2,636 J q= -2,636 J released Conclusion: My hypothesis was correct. The water did in fact drastically rise as the sodium chloride was dissolving in the water. As a result, -2,636 Joules of heat was released from the chemical reaction with water and NaCl. The reaction was an Exothermic Process, because heat was released into the surroundings. The system in this reaction was the NaCl dissolving, and the surrounding being the
Determining the Chemical Formula of a Hydrate Purpose: Find the molecular formula of the hydrate of Copper (II) Sulfate. Materials: * 400 mL beaker * Tongs * Scoopula * Electronic Balance * Glass Rod * Hot Pad * 3 g hydrated copper(II) sulphate * Eye Protection Hypothesis: This lab will demonstrate a dehydration reaction. When the beaker containing hydrated copper (II) sulphate is heated, the water will evaporate, causing the color to change from blue to white, and the mass to decrease. Procedure: 1. Measure & record mass of beaker 2.
Siddharth Rajendran Chemistry HL Urea Dissolution Lab Raw Data:- (Expected Values) Change in Enthalpy: 14 kJ mol-1 Change in Gibbs free energy: 6.86 kJ mol-1 Change in Entropy: 69.5 J mol-1 Molar Mass of Urea: 60.06 g mol-1 Heat Capacity: 4.184 J g-1º Data Table 1: To calculate the Enthalpy change Mass of Urea Tablet (g) (±0.01g) | Volume of Water(mL) (±0.05mL) | Initial Temperature (Cº) (±0.2 Cº) | Final Temperature(Cº) (±0.2 Cº) | 3.04 | 50.0 | 21.3 | 17.4 | Initial Observations:- * There was a decrease in temperature at a fast rate. * The temperature of the solution was slowing down continuously but the rate started decreasing. * The Urea dissolved and the rate was decreasing continuously. * The temperature gradually started to increase after almost the Urea present had dissolved. Data Table 2: Mass, Volume and Temperature during Dissolution of Urea (To calculate Keq) Mass of Urea(g) (±0.01g) | Initial Temperature(Cº) (±0.2 Cº) | Final TemperatureCº) (±0.2 Cº) | Initial Volume(mL)(±0.05 mL) | Final Volume(mL)(±0.05 mL) | 3.76 | 21.4 | 22.9 | 5.02 | 7.14 | Processing Raw Data * Determining the Final temperature of dissolution of Urea in the Styrofoam cup.
- 1 Spectrophotometer. Method to obtain solutions: - Weigh 0.25g of Cu powder in a beaker. - Under fuming cupboard add 1 cm^3 of Nitric Acid in the same beaker. - Pour the Copper Nitrate solution into the 250 cm^3 volumetric flask and dilute it with water until it reaches the line. - Using a pipette pour 5cc, 10cc, 15cc and 20cc of the solution into different volumetric flasks each of 100 cm^3 volume and dilute them until the line mark and
Controlled The controlled variables of the experiment were: A. The volume and concentration of the Hydrochloric acid. B. The concentration of the Sodium Hydroxide. Equipment List * Boiling Tube * 10 cm3 1mol dm-3 Hydrochloric Acid (HCL) * 15 cm3 1mol dm-3 Sodium Hydroxide (NaOH) * pH and Temperature Probes * Data Logger * Measuring Cylinder ‘ * Boiling Tube * Teat Pipette Method * Add 10ml of Hydrochloric acid, measured in a measuring cylinder, into a boiling tube.