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.
Precipitation occurs when the product of the ion concentration exceeds the Ksp.
A) c = 3 x 10^8 m/s
f = 7.15 x 10^14 Hz
c = λ x f (=) λ = 3 x 10^8 / 7.15 x 10^14 = 4.19 x 10^-7 m = 419.6 nm
B) E = h f
H = Planck's constant = 6.63 x 10^-34 J/s
E = 6.63 x 10^-34 x 7.15 x 10^14 = 4.74 x 10^-19 J
Read more on Brainly.com -
brainly.com/question/5760368#readmore
The rate of chemical reactions generally happen <em>faster</em> when the temperature is raised.
This happens because the reactant's molecules move faster when the temperature is raised. The molecules start to bounce around more, increasing the chance for the reaction to happen, or to increase the speed at which the reaction occurs. Hope this helped.
Answer:
C + 2H2 ⇒ CH4
Explanation:
In order to balance a chemical equation you need to make sure that the number of atoms on both sides are equal
C + H2 = CH4
C = 1
H = 2
Products:
C = 1
H = 4
H2 = 2 × 2 = 4
C + 2H2 ⇒ CH4
Hope this helps.
