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.
The number of moles of the magnesium (mg) is 0.00067 mol.
The number of moles of hydrogen gas is 0.0008 mol.
The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.
<h3>
Number of moles of the magnesium (mg)</h3>
The number of moles of the magnesium (mg) is calculated as follows;
number of moles = reacting mass / molar mass
molar mass of magnesium (mg) = 24 g/mol
number of moles = 0.016 g / 24 g/mol = 0.00067 mol.
<h3>Number of moles of hydrogen gas</h3>
PV = nRT
n = PV/RT
Apply Boyle's law to determine the change in volume.
P1V1 = P2V2
V2 = (P1V1)/P2
V2 = (101.39 x 146)/(116.54)
V2 = 127.02 mL
Now determine the number of moles using the following value of ideal constant.
R = 8.314 LkPa/mol.K
n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)
n = 0.0008
<h3>Volume of 1 mole of hydrogen gas at STP</h3>
V = nRT/P
V = (1 x 8.314 x 273) / (101.325)
V = 22.4 L
Learn more about number of moles here: brainly.com/question/13314627
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Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
