Conditions:
Low pressure and low temperature
Low pressure and high temperature
High pressure and low temperature
High pressure and high temperature
<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
Answer:
Iron(III) Oxide
Explanation:
You can tell that this formula is for the molecule Iron(III) oxide because it has two iron atoms and three oxygen atoms.
Fun Fact: There are three main types of iron oxides, with this being one of them.
Hope this helped! :^)
Answer:
D: lose an electron
Explanation:
when an atom loses an electron it's positively charged and when it gain an electron it is negatively charged
When you heated the can with the bit of water inside and you boiled it over a flame, the water turned to vapor (gas) and the pressure in the inside of the can is different from the pressure on the outside of the can. When you placed the can into a ice water beaker or a container, the can shrunk it's size, decreasing it's mass and density. The can shrunk as a result of the inside pressure being equalized with the outside pressure.
The part where you placed it in the ice bath or container was when the water vapor was forced out of the can.