If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
1 answer:
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
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Answer: b} The exact time when an individual atom will decay can be accurately predicted.
c} After each half-life, the amount of radioactive material is reduced by half.
Explanation:
All radioactive decay follows first order kinetics.
Rate law expression for first order kinetics is given by:
where,
k = rate constant
t = time taken for decay process
a = initial amount of the reactant
a - x = amount left after decay process
Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:
Answer:
14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.
Explanation:
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.
During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.
So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.
In this case, being:
- Q=?
- m= 172 g
- L= 84
and replacing in the expression Q = m*L you get:
Q=172 g*84
Q=14,448 J
<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>