<h2>
Hello!</h2>
The answer is:
The temperature will be the same, 37°C.
<h2>
Why?</h2>
Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.
The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.
The law establishes the following equation:
Where,
is the first pressure.
is the first volume.
is the first temperature.
is the second pressure.
is the second volume.
is the second temperature.
Then, we are given the following information:
So, isolating the new temperature and substituting the given information, we have:
Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.
The temperature will be the same, 37°C
Have a nice day!
Answer:
B) -1.00 M/min
Explanation:
Based on the reaction:
2 SO2(g) + O2(g) → 2SO3(g)
<em>2 moles of SO2 are consumed per mole of O2 producing 2 moles SO3</em>
<em />
In the problem, the sulfur dioxide has a rate of change of -2.00M/min. As the moles of oxygen that reacted are half the moles of SO2, its rate of change is halved, that is:
<h3>B) -1.00 M/min
</h3>
Answer:
7 L of H₂.
Explanation:
The balanced equation for the reaction is given below:
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
1 L of O₂ required 2 L of H₂.
Finally, we shall determine the volume of H₂ required to react with 3.5 L of O₂. This can be obtained as follow:
From the balanced equation above,
1 L of O₂ required 2 L of H₂.
Therefore, 3.5 L of O₂ will require
= 3.5 × 2 = 7 L of H₂.
Thus, 7 L of H₂ is required to for the reaction.
D because Carbon and Oxygen form covalent compounds. I wasn't the greatest at chem., but I'm pretty sure this is correct :D let me know if I gave you the right answer.
Answer:
Self-reduction is a reaction in which no external reducing agent is required. It is called self-reducing because copper oxide itself reduced the copper sulfide to give copper and sulfur dioxide.
