Answer:
D. It is limited to situations that involve aqueous solutions or specific compounds.
Explanation:
An Arrhenius acid is a substance that increases the concentration of H3O or H+ when dissolved in water. An Arrhenius base is a substance that increases the concentration of OH- when dissolved in water. These definitions tell us that D is indeed limited to situations that involve aqueous solutions or specific compounds, as aqueous means something that's dissolved in water.
A is wrong because the Bronsted-Lowry interpretation has a wider range of applications. Bronsted-Lowry acids and bases don't even need to be aqueous, so it is not limited to just aqueous solutions. They include any substance that can donate or accept a H+.
B is wrong because A is wrong. A and B basically say the same thing, that the Arrhenius interpretation has a wider range of applications than the Bronsted-Lowry interpretation.
C is wrong because the definition of an Arrhenius base is any substance that increases the concentration of OH-, or hydroxide ions. C completely counters this statement.
Here's photo for proof incase you're doubtful of my answer & explanation. Please click the heart if it helped.
Answer:
Si is the answer I hope this help
Answer:
If you put friction in to skating rink then obvioulsy the skaters wont be able to glide and skate.
Explanation:
friction is power that disturbes the moving power. places like skating rink and slides try to have the least amount of friction.
Answer:
Explanation:
2C₄H₁₀ + 13O₂ = 8 CO₂ + 10H₂O
Change in number of moles Δn = 18 - 15 = + 3 moles .
ΔHo = -2658.3 kJ/mol.
ΔHo = ΔEo+ Δn RT
Δn = 3
For one mole Δn = 1.5
ΔHo = ΔEo+ W
W = Δn RT
= 1.5 x 8.31 x 298
= 3714.5 J
= 3.7 kJ /mole
ΔHo = ΔEo+ W
ΔEo = ΔHo - W
= -2658.3 - 3.7 kJ
= - 2662 kJ .
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
