Answer : 17.12 g
Explanation:![\Delta T =k_b\times m](https://tex.z-dn.net/?f=%5CDelta%20T%20%3Dk_b%5Ctimes%20m)
= elevation in boiling point
= boiling point elevation constant
m= molality
![molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}](https://tex.z-dn.net/?f=molality%3D%5Cfrac%7Bmass%20of%20solute%7D%7Bmolecular%20mass%20of%20solute%5Ctimes%20weight%20of%20the%20solvent%20in%20kg%7D)
given ![\Delta T=6.28^{\circ}C](https://tex.z-dn.net/?f=%5CDelta%20T%3D6.28%5E%7B%5Ccirc%7DC)
Molar mass of solute = 46.0 ![gmol^{-1}](https://tex.z-dn.net/?f=gmol%5E%7B-1%7D)
Weight of the solvent = 150.0 g = 0.15 kg
Putting in the values
![molality=\frac{x}{46gmol^{-1}\times0.15kg}](https://tex.z-dn.net/?f=molality%3D%5Cfrac%7Bx%7D%7B46gmol%5E%7B-1%7D%5Ctimes0.15kg%7D)
![6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}](https://tex.z-dn.net/?f=6.28%20%3D2.53%5E%7B%5Ccirc%7DCkgmol%5E%7B-1%7D%5Cfrac%7Bx%7D%7B46gmol%5E%7B-1%7D%5Ctimes%200.15kg%7D)
x = 17.12 g
Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)
Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)
Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2
Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g
1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au
Volume = Area (of the face) * Length
We want Length:
Length = Volume/Area
L = (518.5 cm^3/0.606 cm^2)
L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)
Water freezes at 0°C, 32°F, and 273°K. The only temperature warmer than the freezing point is 1°C.
In this question, you are given the gasoline density (0.749g/ml) and volume of the gasoline (19.2 gallons). You are asked the mass of the gasoline in pounds. Then you need to change the grams into pounds and the ml into gallons. The calculation would be:
mass of gasoline= density * volume
mass of gasoline= 0.749g/ml * (1 pound/453.592grams) * 3785.41ml/gallon * 19.2 gallon= 120 pounds
<span>58.6934 +/- 0.0002 u</span>