The freezing point depression of water is 1.86°C/m.kg
<span>K2S dissociates: </span>
<span>K2S ↔ 2K+ + S 2- </span>
<span>1mol K2S will produce 3 mol of ions. </span>
<span>Therefore in the freezing point equation: i = 3 </span>
<span>Depression of freezing point = Kf * i* m </span>
<span>Depression of freezing point = 1.86*3*0.195 </span>
<span>Depression of freezing point = 1.088°C </span>
<span>The solution will freeze at - 1.088°C </span>
All elements in their standard states (oxygen<span> gas, solid carbon in the form of graphite, etc.) have a standard </span>enthalpy of formation<span> of </span>zero<span>, as there is no change involved in their </span>formation<span>.</span>
Since Titan orbits roughly along Saturn's equatorial plane, and Titan's tilt relative to the sun is about the same as Saturn's, Titan's seasons are on the same schedule as Saturn's—seasons that last more than seven Earth years, and a year that lasts 29 Earth years.
The answer to the problem is 7/10
Answer:
1.45 x 10⁻² g CO₂
Explanation:
To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).
Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)
Molar Mass (CH₄): 16.043 g/mol
Combustion of Methane:
1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole
= 0.0145 g CO₂
= 1.45 x 10⁻² g CO₂
