<u>Answer and Explanation:</u>
Mercury combines with sulfur as follows -
Hg + S = HgS
Hg = 200,59
S = 32,066 Therefore 1.58 g of Hg will react with -
1.58 multiply with 32,066 divide by 200,96 of sulfur.
= 0.25211 g S
This will form 1.58 + 0.25211 g HgS = 1.83211 g HgS
The amount of S remaining = 1.10 - 0.25211 = 0.84789 g
Both are considered mixtures<span> - that is, they are made up of two or more pure substances. ... A </span>heterogeneous mixture<span> appears to be made of </span>different<span>substances. A </span>solution<span> appears the same throughout. In the fluid phase (gas or liquid, or any combination of those) a </span>solution<span> is transparent (thought not colourless). hope that helps</span>
Explanation:
It is assumed that the particles of an ideal gas have no such attractive forces. The motion of each particle is completely independent of the motion of all other particles. The average kinetic energy of gas particles is dependent upon the temperature of the gas.
<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.
