Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.
Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass = 62.5 g
Molar mass of
= 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:

Consider the given reaction as:

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give ![\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr The conversion of P(torr) to P(atm) is shown below: [tex]P(torr)=\frac {1}{760}\times P(atm)](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%7B2%7D%5Ctimes%201.8382%20mole%20of%20oxygen%20gas.%20%3C%2Fp%3E%3Cp%3EMoles%20of%20oxygen%20gas%20produced%20%3D%200.9191%20mol%3C%2Fp%3E%3Cp%3EGiven%3A%20%3C%2Fp%3E%3Cp%3EPressure%20%3D%20746%20torr%0A%3C%2Fp%3E%3Cp%3EThe%20conversion%20of%20P%28torr%29%20to%20P%28atm%29%20is%20shown%20below%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%28torr%29%3D%5Cfrac%20%7B1%7D%7B760%7D%5Ctimes%20P%28atm%29)
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>
<h2>
Hello!</h2>
The answer is:
The new temperature will be equal to 4 K.

<h2>
Why?</h2>
We are given the volume, the first temperature and the new volume after the gas is compressed. To calculate the new temperature after the gas was compressed, we need to use Charles's Law.
Charles's Law establishes a relationship between the volume and the temperature at a gas while its pressure is constant.
Now, to calculate the new temperature we need to assume that the pressure is kept constant, otherwise, the problem would not have a solution.
From Charle's Law, we have:

So, we are given the following information:

Then, isolating the new temperature and substituting the given information, we have:




Hence, the new temperature will be equal to 4 K.

Have a nice day!
collect this money and use it to finance social projects
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The compound is used in medicine as a source of magnesium ions, which are essential for many cellular activities. Magnesium chloride has also been used as a cathartic and in alloys. To low