Hey there!
I can't be sure my answers are the exact words, but it should be something along the lines of...
The amount of water on Earth is constant, but the form and location of the water changes as it moves through the water cycle.
This means that Earth has always had the same amount of water within in, along with it being the same water the whole time. No new water was introduced to our planet. Some of our water is liquid, some is solid, and some is gas. Some is deep in the soil and some is high up in the atmosphere. Some is in rain and some is in snow.
I hope this helps!
Answer:
Gravity
Explanation:
i dont rlly know where it comes from...
Answer:
THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.
Explanation:
Mass = 3.0 kg = 3 * 1000 = 3000 g
Initial temperature = 10 C
Final temperature = 80 C
Change in temperature = 80 - 10 = 70 C
Specific heat of water = 4.18 J/g C
Heat needed = unknown
Heat is the amount of energy in joules needed to change a gram of water by 1 C.
Heat = mass * specific heat * change in temperature
Heat = 3000 g * 4.18 J/g C * 70 C
Heat = 877 800 Joules
Heat = 877.8 kJ.
The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
