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Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.
<h3>What is the freezing point?</h3>
The temperature at which a liquid solidifies.
Sucrose doesn't show appreciable dissociation in water, so i≈1 in this case. by extension, −7.3°C is the approximate freezing point of this solution.
Learn more about the freezing point here:
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Answer:
True
Explanation:
If you see distilled water, it's a pure substance. That means that there are only water molecules in the liquid. A mixture would be a glass of water with other things dissolved inside, like one of those powders you take if you get sick or Kool-Aid.
Answer:
A reaction that is at equilibrium is not capable of doing any work
Explanation:
Chemical equilibrium is the state of a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. While a reaction is in equilibrium the concentration of the reactants and products are constant.
Chemical equilibrium, a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs. A reversible chemical reaction is one in which the products, as soon as they are formed, react to produce the original reactants. At equilibrium, the two opposing reactions go on at equal rates, or velocities, hence there is no net change in the amounts of substances involved. At this point the reaction may be considered to be completed; i.e., for some specified reaction condition, the maximum conversion of reactants to products has been attained.
Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
