Give the person above me brainless
Answer:
B. 111 J
Explanation:
The change in internal energy is the sum of the heat absorbed and the work done on the system:
ΔU = Q + W
At constant pressure, work is:
W = P ΔV
Given:
P = 0.5 atm = 50662.5 Pa
ΔV = 4 L − 2L = 2 L = 0.002 m³
Plugging in:
W = (50662.5 Pa) (0.002 m³)
W = 101.325 J
Therefore:
ΔU = 10 J + 101.325 J
ΔU = 111.325 J
Rounded to three significant figures, the change in internal energy is 111 J.
1. The reaction for this would be:
Sn + 4 HNO₃ → SnO₂ + 4 NO₂ + 2 H₂O
The first observation would be bubbling of the solution and brown acrid smoke is produced due to the presence of NO₂ gas. Another observation would be the presence of a white solid which is SnO₂.
2. Heating was required to get rid of the H₂O. When all moisture is gone, you weigh the sample. Afterwhich, you further heat it to get ride of the oxygen. By doing this, you would know the individual mass of each element. Then, you can solve for the empirical formula of the oxide of tin.
Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = 
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
![K = \frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
where
K is given as ; 78.2 atm-1.
So, we have:
![78.2=\frac{[0.0530-x]}{[x][x]}](https://tex.z-dn.net/?f=78.2%3D%5Cfrac%7B%5B0.0530-x%5D%7D%7B%5Bx%5D%5Bx%5D%7D)


Using quadratic formula;

where; a = 78.2 ; b = 1 ; c= - 0.0530
=
or 
=
or 
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:

where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???

V = 1.64604
V ≅ 1.65 L