Answer:
The new temperature of the nitrogen gas is 516.8 K or 243.8 C.
Explanation:
Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

In this case:
- P1= 2 atm
- T1= 50 C= 323 K (being 0 C= 273 K)
- P2= 3.2 atm
- T2= ?
Replacing:

Solving:


T2= 516.8 K= 243.8 C
<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>
Answer:
Only the H is unbalanced.
Explanation:
There are 4 H's and 2 of everything else
Answer:
4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;
Explanation:
a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation
Answer:
<h3>The answer is option C</h3>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of liquid = 15 mL
density = 2.5 g/mL
We have
mass = 15 × 2.5
We have the final answer as
<h3>37.5 g</h3>
Hope this helps you
In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.
Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.
For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.
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