Answer:
V₂ = 106.5 mL
Explanation:
Given data:
Initial volume =200 mL
Initial pressure = 2 atm
Initial temperature = 35 °C (35 +273 = 308 K)
Final temperature = 55°C (55+273 = 328 K)
Final volume = ?
Final pressure = 4 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2 atm ×200 mL × 328 K / 308 K ×4 atm
V₂ = 131200 atm .mL. K / 1232 K.atm
V₂ = 106.5 mL
Hi there, the correct answer to this would be reactivity because color, melting point, and density are all examples of physical properties.
is the type of orbital hybridization of a central atom that has one lone pair and bonds to four other atoms.
<h3>What is
orbital hybridization?</h3>
In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.
For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.
Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.
Learn more about hybridization
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Answer:
A Type of Drink
Explanation:
A controlled variable remains constant throughout the experiment.
In such experiment, you'd test the volume of one single caffeinated drink. You'd have to use the same type of drink every trial.
Answer:
ΔS° = - 47.2 J/mol.K
Explanation:
ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10
∴ S°mH2O(l) = 69.9 J/mol.K
∴ S°mP4O10 = 231 J/mol.K
∴ S°mH3PO4 = 150.8 J/mol.K
⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231
⇒ ΔS° = - 47.2 J/mol.K
