Answer:
is the change in enthalpy associated with the combustion of 530 g of methane.
Explanation:

Mass of methane burnt = 530 g
Moles of methane burnt = 
Energy released on combustion of 1 mole of methane = -890.8 kJ/mol
Energy released on combustion of 33.125 moles of methane :


is the change in enthalpy associated with the combustion of 530 g of methane.
MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
Explanation:
It is given that energy to transfer one water molecule is
J/molecule
As it is known that in 1 mole there are
atoms.
So, energy in 1 mole =
J/mol
= 13.3 kJ/mol
As, 
Putting the given values in the above formula as follows.


= 0.08377
= 1.213 =
= 
= 
=
water molecules per oil molecule
Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is
.
Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous
Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m