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Answer:
ΔH = 57.04 Kj/mole H₂O
Explanation:
60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)
=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)
=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)
=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat
ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O
=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)
ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O
Answer:
Answer of question a is 345J.
Explanation:
In question a following is given in data:
-mass of iron (m) = 10.0 g
-temperature (ΔT) = final temperature- initial temperature= 100-25= 75 degree Celsius
-Specific Heat capacity of iron (c)= 0.46J/g°C.
Heat (Q)=?
Solution:
Formula for Heat is :
Q=m x c x ΔT
Q= 10 x 0.46 x 75
Q= 345 J.
so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.
- From the above formula all other questions can easily be solved from the same procedure.
