Answer:
the enthalpy change for this reaction is -57.7 kJ/mol
Explanation:
Given:
HB₂O₃ = -1272.8 kJ/mol
HCOCl₂ = -218.8 kJ/mol
HBCl₃ = -403.8 kJ/mol
HCO₂ = -393.5 kJ/mol
Those are all standard enthalpies
Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?
The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows
B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂
Where the compounds on the left are the reactants and the compounds on the right are the products
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants
Please be careful with the signs.
Answer:
Molarity = 1.28
Explanation:
M = molar concentration
n = moles of solute
v = liters of solution
Molarity = moles of solute / litres of solution.
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reference - https://www.wikihow.com
The mercury (Hg) has 80 electrons. The electrons are distributed in the s, p, d and f orbitals. The electronic configuration of Hg can be shown as-. We can see that there are two filled d-orbitals 3d and 4d. There is only one filled f orbital which is 4f. The electron cloud of the s, p, d and f are different due to their different screening and penetration effect towards the nucleus. The electron cloud of f-orbital is most diffuse than the other three. The outermost electrons of mercury is , which is its valence electrons i.e. (+2).
Answer:
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<u>Answer:</u> The moles of oxygen and carbon dioxide in air is and respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of atmosphere =
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:
We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air =
Moles of carbon dioxide in air =
Hence, the moles of oxygen and carbon dioxide in air is and respectively