The action that would most likely lead to. Decrease in volume is decrease in temperature.
Answer:
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Explanation:
Equating coefficients, you get ...
aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃
For Ba: 3a = d
For P: 2a = 4c
For O: 8a +2b = 10c +3d
For Si: b = d
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Expressing everything in terms of b and c, we get ...
d = b
a = b/3 = 2c
From the second, b = 6c, so we have ...
a = 2c
b = 6c
c = c
d = 6c
And we can write the equation with c=1 as ...
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
Answer:
34.9 g/mol is the molar mass for this solute
Explanation:
Formula for boiling point elevation: ΔT = Kb . m . i
ΔT = Temperatures 's difference between pure solvent and solution → 0.899°C
Kb = Ebullioscopic constant → 0.511°C/m
m = molality (moles of solute/1kg of solvent)
i = 2 → The solute is a strong electrolyte that ionizes into 2 ions
For example: AB ⇒ A⁺ + B⁻
Let's replace → 0.899°C = 0.511 °C/m . m . 2
0.899°C / 0.511 m/°C . 2 = m → 0.879 molal
This moles corresponds to 1 kg of solvent. Let's determine the molar mass
Molar mass (g/mol) → 30.76 g / 0.879 mol = 34.9 g/mol
