Baloon with 3 moles og oxygen at 1 atm.The temperature of the balloon is <u>4 Kelvin</u>.
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Using the ideal gas equation:-
Given;
P₁ = 1 atm
V₁ = 100 L
n = 3
r = 8.314
T = PV/nR
= 1 × 100 / 3 × 8.314
= 4 K
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Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
And the undergoing chemical reaction:
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
Next, the moles of magnesium chloride consumed by the sodium fluoride:
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
![[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M](https://tex.z-dn.net/?f=%5BMg%5E%7B2%2B%7D%5D%3D%5Cfrac%7B3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D3.75x10%5E%7B-4%7DM)
![[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M](https://tex.z-dn.net/?f=%5BF%5E-%5D%3D%5Cfrac%7B2%2A3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D7.5x10%5E%7B-4%7DM)
Thereby, the reaction quotient is:
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Answer:
441.28 g Oxygen
Explanation:
- The combustion of hydrogen gives water as the product.
- The equation for the reaction is;
2H₂(g) + O₂(g) → 2H₂O(l)
Mass of hydrogen = 55.6 g
Number of moles of hydrogen
Moles = Mass/Molar mass
= 55.6 g ÷ 2.016 g/mol
= 27.8 moles
The mole ratio of Hydrogen to Oxygen is 2:1
Therefore;
Number of moles of oxygen = 27.5794 moles ÷ 2
= 13.790 moles
Mass of oxygen gas will therefore be;
Mass = Number of moles × Molar mass
Molar mass of oxygen gas is 32 g/mol
Mass = 13.790 moles × 32 g/mol
<h3> = 441.28 g</h3><h3>Alternatively:</h3>
Mass of hydrogen + mass of oxygen = Mass of water
Therefore;
Mass of oxygen = Mass of water - mass of hydrogen
= 497 g - 55.6 g
<h3> = 441.4 g </h3>
H20. 2 of hydrogen and oxygen
