Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
Answer:
The Answer is 'D'
Explanation:
The diagram on the down side shows the behavior of the particles of a liquid so I suppose it is the ocean. While the top diagram shows the behavior of the particles of a gas so I am sure it's the air. Therefore I chose the last diagram because it describes exactly how you wanted in the question, which is the Ocean's water evaporating to become gas or the 'air' as we say
<em>Thank</em><em> </em><em>you</em><em> </em><em>and</em><em> </em><em>I</em><em> </em><em>hope</em><em> </em><em>you</em><em> </em><em>like this</em><em> </em><em>answer</em><em>! </em>
To begin calculating, there is one thing you need to remember :
Then we have
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>
I'm sure it will help.
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
Sorry, the correct answer is A. 3.4 x 10 - 24
Hope I helped ; )
