Tutor profile: Bob W.
Subject: Basic Math
Write the equation for the line that passes through the points (-1,-2) and (2,4) using the slope-intercept form, Y = mX + b.
1. Determine the slope of the line, m. m = change in Y/change in X m = [(4) – (-2)]/[(2) – (-1)] m = 6/3 m = 2 2. Determine b by using m, X, & Y, using either point. Y = mX + b Using (2,4): (4) = 2(2) + b 4 = 4 + b b = 0 Using (-1,-2) (-2) = 2(-1) + b -2 = -2 + b b = 0 3. Substitute m and b back into the equation. Y = mX + b Y = 2X + 0 Y = 2X
Using a eudiometer, 43.68 mL of butane was collected by water displacement from a lighter. The lighter and water were set out over night to reach a constant temperature. A student collected the following data in order to experimentally determine the molar mass of butane. Data: Water temperature: 21.3 degrees C Mass of lighter prior to gas collection: 28.9550 g Mass of lighter after gas collection: 28.8525 g Atmospheric pressure: 741.08 mm Hg Using the data above, calculate the molar mass of butane.
1. Modify the ideal gas law equation to include Molar Mass by substituting mass/Molar Mass in for the number of moles, n. PV = nRT n = m/MM PV = mRT/MM MM = mRT/PV 2. Determine the values for m, T, V, and P that have units compatible with the universal gas constant R (R=0.08206 L x atm/(mol x K). a. Mass of the butane, m in grams. m = mass of lighter at the start – mass of lighter at the end m = 28.9550 g – 28.8525 g m = 0.1025 g b. Temperature of the butane, T in Kelvins. T = 21.3 degrees C + 273.15 T = 294.5 K c. Volume of the butane, V in liters. V = 43.68 mL x 1 L/1000.0 mL V = 0.04368 L d. Pressure of the dry butane, P in atmospheres. P = Pressure of the gas mixture – Vapor pressure of water @ 21 degrees C P = 741.08 mm Hg – 18.7 mm Hg P = 722.38 mm Hg P = 722.38 mm Hg x 1 atm/760.0 mm Hg P= 0.9505 atm 3. Insert m, T, P, and V into the modified ideal gas law equation. MM = mRT/PV MM = (0.1025 g)(0.08206 L x atm/(mol x K))(294.5 K)/(0.9505 atm x 0.04368 L) MM = 59.67 g/mol
Subject: Basic Chemistry
A sample of gas is placed in a sealed container with a moveable piston. If the piston is moved so the volume of the gas doubles and the temperature remains the same, 1. The pressure will double. 2. The pressure will be cut in half. 3. The number of molecules will double. 4. The number of molecules will be cut in half.
The correct answer is choice 2. Since the container is sealed, the number of gas molecules will not change so choices 3 & 4 are incorrect. The pressure of a gas is inversely proportional to the volume when the temperature and number of gas molecules stay the same. In an inverse proportion, when one variable increases, the other must decrease so the product of the two remains constant.
needs and Bob will reply soon.