The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or 1486.788 mm Hg.
<h3>The ideal gas law</h3>
According to the ideal gas law, the product of the pressure and volume of a gas is a constant.
This can be mathematically expressed as:
pv = nRT
Where:
p = pressure of the gas
v = volume
n = number of moles
R = Rydberg constant (0.08206 L•atm•mol-1K)
T = temperature.
In this case:
p is what we are looking for.
v = 5.00 L
n = 0.400 moles
T = 25 + 273
= 298 K
Now, let's make p the subject of the formula of the equation.
p = nRT/v
= 0.400 x 0.08206 x 298/5
= 1.9563 atm
Recall that: 1 atm = 760 mm Hg
Thus:
1.9563 atm = 1.9563 x 760 mm Hg
= 1486.788 mm Hg
In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.
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I think answer is
C. Something that can be observed or measured while changing the identity of the substance
B. At the equivalence point of a titration of the [H+] concentration is equal to 7.
<h3>What is equivalence point of a titration?</h3>
The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.
At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.
At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;
H⁺ + OH⁻ → H₂O
The pH of resulting solution is 7.0 (neutral).
Thus, the pH at the equivalence point for this titration will always be 7.0.
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Answer:
42.65g
Explanation:
Given parameters:
Mass of K = 4g
Unknown: Mass of KCl
Solution:
Complete equation of the reaction:
2K + Cl₂ → 2KCl
To solve this problem, we know that the reactant in short supply is potassium K and this dictates the amount of products that would be formed. The chlorine gas is in excess and we can't use it to determine the amount of product that would form.
Now, we work from the known to the unknown. Since we know the mass of K given in the reaction, we can simply find the molar relationship between the reacting potassium and the product. We simply convert the mass to mole and compare to the product. From there we can find the mass of KCl that would be produced.
Calculating number of moles of K
Number of moles = 
Number of moles of K =
= 0.103mol
From the given reaction equation:
2 moles of K will produce 2 moles of KCl
Therefore 0.103mol of K will produce 0.103mol of KCl
To find the mass of KCl produced,
Mass of KCl = number of moles of KCl x molar mass
Molar mass of KCl = 39 + 35.5 = 74.5gmol⁻¹
Mass of KCl = 0.103 x 74.5 = 42.65g