1.you have an unknown amount of water vapor floating in a sealed container at 126℃. after 30 min the container is cooled to -26℃, and a solid sample of water is found at 2.5g. what amount of energy was used for this process to occur?

7.8 kJ. The process used 7.8 kJ.

The container is sealed, so

mass of water vapour = mass of water = mass of ice.

∴ Heat required = heat to cool vapour + heat to condense vapour to water + heat to cool water + heat to freeze water + heat to cool ice

q = q_1 + q_2 + q_3 + q_4 + q_5

Step 1. Calculate q_1

q_1 = mC_1ΔT_1

m = 2.5 g; C_1 = 1.996 J.°C^(-1)g^(-1)

ΔT_1 = T_f – T_i  = 100 °C – 126 °C = -26 °C

q_1 = 2.5 g × 1.996 J.°C^(-1)g^(-1) × (-26°C) = -130 J

The negative sign shows that heat is removed from the system.

Step 2. Calculate q_2

q_2 = mΔ_cH

Δ_cH = -2257 J·g^(-1)

q_2 = 2.5 g × [-2257 J·g^(-1)] = -5640 J

Step 3. Calculate q_3

q_3 = mC_3ΔT_3

C_3 = 4.182 J.°C^(-1)g^(-1); ΔT_3 = T_f – T_i  = 0 °C – 100 °C = -100 °C

q_3 = 2.5 g × 4.182 J.°C^(-1)g^(-1) × (-100 °C) = -1050 J

Step 4. Calculate q_4

q_4 = mΔ_cH

Δ_cH = -334 J·g^(-1)

q_4 = 2.5 g × [-334 J·g^(-1)] = -835 J

Step 5. Calculate q_5

q_5 = mC_5ΔT_5

C_5 = 2.090 J.°C^(-1)g^(-1); ΔT_3 = T_f – T_i  = -26 °C –0 °C = -26 °C

q_5 = 2.5 g × 2.090 J.°C^(-1)g^(-1) × (-26 °C) = -136 J

Step 6. Calculate the total heat involved

q = q_1 + q_2 + q_3 + q_4 + q_5

= (-130 – 5640 – 1050 – 835 – 136) J = -7800 J = -7.8 kJ

The process removed 7.8 kJ.

answer is: naturally occurring solid, inorganic compound or element is mineral.

minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

for example magnetite is a mineral of iron oxide (fe₂o₃).

calcite is a carbonate mineral and the most stable polymorph of calcium carbonate (caco₃).

dolomite is an   carbonate mineral composed of calcium magnesium carbonate (camg(co₃)₂).