Answer:
0.521 moles still present in the container.
Explanation:
It is possible to answer this question by using the general gas law, that is:
PV = nRT
<em>Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)</em>
Replacing with values of the initial conditions of the container, its volume is:
V = nRT / P
V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm
V = 12.9L
When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and <em>the volume of the container still constant. </em>Again, using general gas law:
PV / RT = n
0.998atm*12.9L / 0.082atmL/molK*301.25K = n
0.521 moles = n
<h3>0.521 moles still present in the container.</h3>
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Answer:
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Explanation:
Step 1: Data given
Mass of Al2O3 = 3.50 grams
Molar mass of Al2O3 = 101.96 g/mol
Number of Avogadro = 6.022 * 10^23 /mol
Step 2: Calculate moles Al2O3
Moles Al2O3 = mass Al2O3 / molar mass Al2O3
Moles Al2O3 = 3.50 grams / 101.96 g/mol
Moles Al2O3 = 0.0343 moles
Step 3: Calculate moles Aluminium
In 1 mol Al2O3 we have 2 moles Al
in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al
Step 4: Calculate aluminium atoms
Aluminium atoms = moles aluminium * Number of Avogadro
Aluminium atoms = 0.0686 * 6.022 * 10^23
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
a scale-model mound made of the same materials that make the real hill
Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.
If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3
The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M
