Answer:
T final = 80°C
Explanation:
∴ Q = 18000 cal
∴ m H2O = 300 g
∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K
∴ T1 = 20°C = 293 K
∴ T2 = ?
⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)
⇒ (18000 cal)/(300 cal/K) = T2 - 293 K
⇒ T2 = 293 K + 60 K
⇒ T2 = 353 K (80°C)
I think this is alloys...................
Answer:
2Mg + 4HCl (aq) --> 2 MgCl2 (aq) + H2
Explanation:
balanced...
The two s Orbital electrons and one d orbital electron, are the electrons that are lost by an atom of Iron when it forms the Fe3 + ion.
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.
