Line 1: straight horizontal line
Line 2: straight line at a slope
Line 3: exponential growth curve
Line 4: the topmost curve (the one that initially increases but then starts levels out)
Answer:
Mole fraction of solute is 0.0462
Explanation:
To solve this we use the colligative property of lowering vapor pressure.
First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr
Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.
Formula for lowering vapor pressure is:
ΔP = P° . Xm
Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)
Xm = mole fraction
24.9 mmHg - 23.8 mmHg = 23mmHg . Xm
Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg
Xm = 0.0462
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer:
TRUE
Explanation:
Production of Hydrocarbons from Natural Gas is as stated below:
Natural gas liquids include propane, butane, pentane, hexane, and heptane, but not methane and not <u>always</u> ethane, (<em>may include it </em><em><u>sometimes</u></em><em>.</em>) s<em>ince these hydrocarbons need refrigeration to be liquefied.</em>
Answer:
Final temperature of calorimeter is 25.36^{0}\textrm{C}
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.840 g of anethole = 
of anethole = 0.00567 moles of anethole
1 mol of anethole releases 5539 kJ of heat upon combustion
So, 0.00567 moles of anethole release 
of heat or 31.41 kJ of heat
6.60 kJ of heat increases 
temperature of calorimeter.
So, 31.41 kJ of heat increases 
or 
temperature of calorimeter
So, the final temperature of calorimeter = 
