Answer:
H+/H3O , H2O
Explanation:
The ability to be a proton donor is the Bronsted-Lowry definition of acids. The Lewis definition of an acid is an electron pair acceptor, which covers molecules liKE BF3
The ability to accept a pair of electrons is what is common to all acids, not the ability to be a proton donor.
All acid solutions contain hydronium ions (H3O+), hydroxide ions (OH-) and water molecules. Each different acid solution will then have an anion that is exclusive to that acid. For example, hydrochloric acid solution will contain all of the above and chloride ions (Cl-).
All acids contain the acidic substance dissolved in water. Water naturally dissociates to a small amount, creating hydronium and hydroxide ions. But most of the water remains as water molecules.
Then when we add an acid, like HCl, the oxygen on the water attracts the hydrogen from the HCl. The electrons in the covalent bond remain with the chlorine, giving it a negative charge and thus it becomes the chloride ion (Cl-). The hydrogen now has a positive charge and as said before, is attracted to the water (specifically the lone pair of electrons on the oxygen) to create hydronium ions.
This creates extra hydronium ions, making the solution acidic. But remember, there are still water molecules, hydroxide ions and the negative ion all in solution for all acids.
Answer:
110.8 ºC
Explanation:
To solve this problem we will make use of the Clausius-Clayperon equation:
lnP = - ΔHºvap/RT + C
where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.
Now this equation has a form y = mx + b where
y = lnP
x = 1/T
m = -ΔHºvap/R
Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC
Thus what we have to do is find the equation of the best fit for this data using a software as excel or your calculator.
T ( K) 1/T ln P
377 0.002653 5.9915
384 0.002604 6.2115
390 0.002564 6.3969
395 0.002532 6.5511
398 0.002513 6.6333
The best line has a fit:
y = -4609.5 x + 18.218
with R² = 0.9998
Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.
ln(496) = -4609.5(1/Tb) + 18.218
6.2066 = -4609.5(1/Tb) +18.218
⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261
Tb = 383.76 K = (383.76 -273)K = 110.8 ºC
Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.
Perhaps that is the reason the question was flagged.
The percent A by mass for substance AB =<u> 75%</u>
<h3>Further explanation</h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements
Empirical formula is the mole ratio of compounds forming elements.
From Substance AB₂ is 60.0% A by mass.
Let's say that AB₂ mass = 100 gram, then
mass A = 60 gram
mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram
In compound AB:
Total mass = mass A + mass B
Total mass = 60 + 20 grams = 80 grams
Then the percentage of compound A = (60: 80) = 75%
<h3>Learn more</h3>
Grams of KO₂ needed to form O₂
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Keywords : percent mass, substance
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First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
<span>When an ice cube is placed on a kitchen counter, heat will flow from the ice cube to the counter, causing the molecules in the counter to move more slowly. The molecules of the counter move more slowly because the heat transferred to them from the ice has reduced their kinetic energy.</span>