Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
<h2>Work done = mgh </h2>
Explanation:
- In this case, while lifting the book we are working against the force of gravity.
Using the Newton's laws, we can find the force F required for lifting the book having mass (m) and acceleration due to gravity (g) that is ;
and, the change in the position of the book that is Δx (Height)
→ Δx = Final position - Initial position
which is only the height, then the amount of work done will be calculated by :
W= mgh
m = Mass of the Body
g = Acceleration due to Gravity
h = Height of Body being displaced
Explanation:
(1) The nucleus is positive and the electron cloud is positive.
(2) The nucleus is positive and the electron cloud is negative.
(3) The nucleus is negative and the electron cloud is positive.
(4) The nucleus is negative and the electron cloud is negative
Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
