<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.
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
An atom that has fewer neurons than protons and more electrons than protons is an ANION
An increased temperature would result in an increased rate of dissolution of a solid solute.
Answer:
Amount of mercury is 1.0*10⁻⁵ g
Explanation:
<u>Given:</u>
Mercury content of stream = 0.68 ppb
volume of water = 15.0 L
Density of water = 0.998 g/L
<u>To determine:</u>
Amount of mercury in 15.0 L of water
<u>Calculation:</u>
where 1 μg (micro gram) = 10⁻⁶ g
0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water
Therefore, the amount of mercury in 15.0 L water would be:
