Oil<span> and </span>water<span> are two liquids that are </span>immiscible<span>, meaning they will not mix together. Liquids tend to be </span>immiscible<span> when the force of attraction between the molecules of the same liquid is greater than the force of attraction between the two different liquids.</span>
The answer would be B imo
C and E are wrong right off the bat.
Answer:
- 1273.02 kJ.
Explanation:
This problem can be solved using Hess's Law.
Hess's Law states that <em>regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.</em>
- We should modify the given 3 equations to obtain the proposed reaction:
<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>
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- We should multiply the first equation by (6) and also multiply its ΔH by (6):
6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,
- Also, we should multiply the second equation and its ΔH by (6):
6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.
- Finally, we should reverse the first equation and multiply its ΔH by (- 1):
6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.
- By summing the three equations, we cam get the proposed reaction:
<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>
<em></em>
- And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:
<em>∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ =</em> (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = <em>- 1273.02 kJ.</em>
From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.
From the information we have;
Volume of the damp air = 1 L
Pressure of the damp air = 741.0 torr or 0.975 atm
Temperature of the gas = 20 oC + 273 = 293 K
R = 0.082 atm LK-1mol-1
Number of moles = ?
n =PV/RT
n = 0.975 × 1/0.082 × 293
n = 0.041 moles
Volume of water vapor = 1 L
Temperature of water = -10 oC + 273 = 263 K
Pressure of the gas = 607.1 torr or 0.799 atm
R = 0.082 atm LK-1mol-1
n= PV/RT
n = 0.799 × 1/ 0.082 × 263
n = 0.037 moles
Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles
If 1 mole = 6.02 × 10^23 molecules
0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole
= 2.41 × 10^21 molecules
Learn more: brainly.com/question/2510654
Answer:
An open system
Explanation:
An open system is a system in which both matter and energy are exchanged freely between the system and the surroundings.
An example is a pot of water boiling on the stove. The surroundings (the stove) can supply heat energy to the water and the water can escape into the atmosphere.
A <em>closed system</em> is a system in which energy but not matter is exchanged freely between the system and the surroundings.
An example is a pressure cooker on the stove. The surroundings (the stove) can supply heat energy to the food inside, but no matter can escape through the closed lid.
An <em>isolated system</em> is a system in which neither energy nor matter can be exchanged between the system and the surroundings.
An example is a thermos of hot soup. The cap prevents matter from escaping and the shiny interior reflects heat back into the soup.