We can calculate for temperature by assuming the equation
for ideal gas law:
P V = n R T
Where,
P = pressure = 1.80 atm
V = volume = 18.2 L
n = number of moles = 1.20 moles
R = gas constant = 0.08205746 L atm / mol K
Substituting to the given equation:
T = P V / n R
T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm /
mol K)
T = 332.70 K
We can convert K unit to ˚C unit by subtracting 273.15
to Kelvin, therefore
T = 59.55 ˚<span>C</span>
The density of an ordinay rock is close to 3 g/cm^3 wihle the density of the paper clips is close to 8 g/cm^3 (the density of steel), then equal apparent volumes (same box) will contain different mass, being of course the mass of the box with paper clips much higher than that of the box with rocks.
ANSWER
EXPLANATION
Given that
The energy released by the system is 12.4J
Work done on the surrounding is 4.2J
Follow the steps below to find the change in energy
In the given data, energy is said to be released to the surroundings
Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative
Step 1; Write the formula for calculating change in energy
![\Delta E\text{ }=\text{ q }+\text{ w}](https://tex.z-dn.net/?f=%5CDelta%20E%5Ctext%7B%20%7D%3D%5Ctext%7B%20q%20%7D%2B%5Ctext%7B%20w%7D)
Since heat is released to the surrounding, then q = -12J
Recall, that work done by the system on the surroundings is always negative
Hence, w = -4.2J
Step 2; Substitute the given data into the formula in step 1
![\begin{gathered} \text{ }\Delta E\text{ = q + w} \\ \text{ }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{ }\Delta E\text{ = -12.4 - 4.2} \\ \text{ }\Delta E\text{ }=\text{ -16.6J} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7B%20%20%7D%5CDelta%20E%5Ctext%7B%20%3D%20q%20%2B%20w%7D%20%5C%5C%20%5Ctext%7B%20%20%7D%5CDelta%20E%5Ctext%7B%20%7D%3D%5Ctext%7B%20-12.4%20%7D%2B%5Ctext%7B%20%5Clparen-4.2%5Crparen%7D%20%5C%5C%20%5Ctext%7B%20%20%7D%5CDelta%20E%5Ctext%7B%20%3D%20-12.4%20-%204.2%7D%20%5C%5C%20%5Ctext%7B%20%20%7D%5CDelta%20E%5Ctext%7B%20%7D%3D%5Ctext%7B%20-16.6J%7D%20%5Cend%7Bgathered%7D)
Therefore, the change i
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>
m 9CH₃COOH: 60u×9 = <u>540u</u>
<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g
</u></span>