Answer: Hsub=Hfus+Hvap
Explanation:
The molar heat of vaporization measured in kilojoules per mole, or kJ/mol is the energy needed to make vapor one mole of a liquid. .
The molar heat of sublimation measured in kilojoules per mole, or kJ/mol is the energy needed to sublime one mole of a solid,
the molar heat of fusion measured in kilojoules per mole, or kJ/mol is the energy needed to melt one mole of a solid.
Hess law helps to explain the relationship in physical chemistry stating that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps.
In this context Hess’s law helps to see the several steps involved as the heat of sublimation energy is equal to the sum of vaporization energy and fusion energy.
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>
<u>Answer:</u>
The primary product of the photosystem I is NADPH.
<u>Explanation:
</u>
NADPH is the acronym of Nicotinamide adenine Dinucleotide Phosphate. This NADPH plays a vital role in many chemical reactions that take place during photosynthesis process. NADPH is released as a product in Photosystem I.
In the photosystem I process, the molecule present will absorb sunlight energy and transfers it to electrons to produce NADPH. This NADPH molecule is used as fuel for the chemical process that occurs during the 2nd stage of the photosynthesis process. Its molecular formula is
. It acts as a fundamental metabolite.
I think you mean:
".. reacts with oxygen gas to produce lead (II) oxide and 'sulfur' dioxide..."
The balanced chemical reaction would be
2PbS + 3O2 --> 2PbO + 2SO2
The stiochiometric ratio is this: 3 moles of O2 produces 2 moles of lead (II) oxide. So,
0.500 moles (2/3) = 0.333 moles
Therefore, 0.333 moles of PbO is produced when 0.500 moles of O2 reacts with PbS.