Answer: 1 mol C8H18 produces 9 mol H2O
Explanation: reaction : C8H18 + 12.5 O2 ⇒ 8 CO2 + 9 H2O
Answer:
The solution is 4.93 molal
Explanation:
We assume that solution is aqueous.
We need to apply the colligative property of elevation of boiling point:
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
110.1°C - 100°C = 10.1°C
Kb = 0.512 °C/m
m = molality → unknown
i = Van't Hoff factor (numbers of ions dissolved)
We assume 100 % dissociation: CoCl₃ → Co³⁺ + 3Cl⁻ i = 4
We have 1 mol of Co³⁺ + 3 moles of chlorides
We replace data → 10.1°C = 0.512°C/m . m . 4
10.1°C / (0.512 m/°C . 4) = m → 4.93
Answer:
![V_2=54.0ft^3](https://tex.z-dn.net/?f=V_2%3D54.0ft%5E3)
Explanation:
Hello,
In this case, we can use the Avogadro's law in order to understand the mole-volume behavior as a directly proportional relationship:
![\frac{V_1}{n_1} =\frac{V_2}{n_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7Bn_1%7D%20%3D%5Cfrac%7BV_2%7D%7Bn_2%7D)
Nonetheless, here we are not given neither the identity of the gas nor its molar or atomic mass, for that reason we can apply the aforementioned equation in terms of mass as we are talking about the same gas:
![\frac{V_1}{m_1} =\frac{V_2}{m_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7Bm_1%7D%20%3D%5Cfrac%7BV_2%7D%7Bm_2%7D)
In such a way, the new volume turns out:
![V_2=\frac{V_1m_2}{m_1} =\frac{174ft^3*13.6lbs}{43.8lbs} \\\\V_2=54.0ft^3](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BV_1m_2%7D%7Bm_1%7D%20%3D%5Cfrac%7B174ft%5E3%2A13.6lbs%7D%7B43.8lbs%7D%20%5C%5C%5C%5CV_2%3D54.0ft%5E3)
Best regards.
Answer:
have you finished this yet
Explanation:
Answer:
they're close to filling their outer shell, fulfilling the octet rule
Explanation: