The warmer road surface at the end of a sunny day is the black road because during the day it absorbed more radiation (sunlight) than the withe one
Answer:
Explanation:
Hello,
In this case, we can compute the change in the solution enthalpy by using the following formula:
Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:
It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.
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Answer:
bonding molecular orbital is lower in energy
antibonding molecular orbital is higher in energy
Explanation:
Electrons in bonding molecular orbitals help to hold the positively charged nuclei together, and they are always lower in energy than the original atomic orbitals.
Electrons in antibonding molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. They are always higher in energy than the parent atomic orbitals.
Answer:
This solution is quite lengthy
Total system = nRT
n was solved to be 0.02575
nH20 = 0.2x0.02575
= 0.00515
Nair = 0.0206
PH20 = 0.19999
Pair = 1-0.19999
= 0.80001
At 15⁰c
Pair = 0.4786atm
I used antoine's equation to get pressure
The pressure = 0.50
2. Moles of water vapor = 0.0007084
Moles of condensed water = 0.0044416
Grams of condensed water = 0.07994
Please refer to attachment. All solution is in there.
The reaction of Phosphorous Pentaoxide with water yield Phosphoric Acid as shown below,
P₄O₁₀ + 6 H₂O → 4 H₃PO₄
According to balance equation,
283.88 g (1 mole) P₄O₁₀ requires = 108 g (6 mole) of H₂O
So,
100 g P₄O₁₀ will require = X g of H₂O
Solving for X,
X = (100 g × 108 g) ÷ 283.88 g
X = 38.04 g of H₂O
So, 100 g P₄O₁₀ requires 38.04 g of H₂O, while we are provided with 200 g of H₂O which means that water is in excess and P₄O₁₀ is limiting reagent. Therefore, P₄O₁₀ will control the yield of H₃PO₄. So,
As,
283.88 g (1 mole) P₄O₁₀ produced = 391.96 g (4 mole) of H₃PO₄
So,
100 g P₄O₁₀ will produce = X g of H₃PO₄
Solving for X,
X = (100 g × 391.96 g) ÷ 283.88 g
X = 138.07 g of H₃PO₄
Result:
Theoretical Yield of this reaction is 138.07 g.
