Answer:
The answer to your question is Aluminum
Explanation:
We need to warm from 22°C to 85°C
a) 45 g of water
b) 200 g of aluminum Cp = 0.905 J/cal°C
a) Water
Q = mCpΔT
Q = (45)(1)(85 - 22) = 45(63) = 2835 cal
b) Aluminum
Q = (200)(0.905)(85 - 22) = 114030 cal
Answer:1
Explanation: friction is showing us how in the past
The given chemical reaction is:
The standard heats of formation of 
are:
Δ
) = -285.8kJ/mol[/tex]
Δ
) = -187.6 kJ/mol[/tex]
Calculating the change in heat:
Δ
=∑ΔH
-∑ΔH
= [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]
= -196.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol
Explanation:
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. pH+pOH=14⇒pH=14-3=
