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.
More on the ideal gas law can be found here: brainly.com/question/28257995
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D. All of the answers are true
Answer:
+3
Explanation:
Chlorine is anion with a -1 charge. But they are three chlorine atoms.
-1 * 3 = -3
So they have a -3 charge.
So to balance the compound, the nickel has to be a cation with a +3 charge.
-3 + 3 = 0
Furthermore, a chemical bond always has a 0 charge. Remember that.
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Answer:
a) 965,1 lbf
b) 4,5 kg
c) 1,33 * 10^6 dynes
Explanation:
Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.
Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.
w=mg
In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).
To solve this problem we'll use the following conversion factors:
1 lbf = 1 lbm*ft/s²
1 N = 1 kg*m/s²
1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes
1 ton = 907,18 kg
1 k = 1000 gr
a) m = 30 lbm
b) w = 44 N
First, we clear m of the weight equation and then we replace our data.
c) m = 15 ton
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.