1) Molarity
M = n / V
n: number of moles of solute
V: volume of the solution in liters
n = mass / molar mass = 0.000333 g / 332.32 g / mol = 1*10 ^ - 6 moles
V = 225 ml * 1 liter / 1000 ml = 0.225 liter
M = 10^-6 mol / 0.225 liter = 0.00000444 M
2) ppm
ppm = parts per million
grams of solute: 0.000333 g
grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g
ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm
<u>Given:</u>
Mass of ice = mass of water = 5.50 kg = 5500 g
Temperature of ice = -20 C
Temperature of water = 75 C
<u>To determine:</u>
Mass of propane required
<u>Explanation:</u>
Heat required to change from ice to water under the specified conditions is:-
q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)
= m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)
= 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ
The enthalpy change for the combustion of propane is -2220 kJ/mol
Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane
Molar mass of propane = 44 g/mol
Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g
Ans: 75.15 grams of propane must be combusted.
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Answer:
By definition, there are 6.022×1023 such molecules, or NA such molecules in ONE mole of water. And thus in such a quantity there are NA oxygen atoms, and 2×NA hydrogen atoms...and the mass associated with this numerical quantity of water molecules is approx. 18⋅g ...
