The question I need help with is #5 on the attached image. How do I calculate the change in heat using the given information in #5? I have tried multiple times and cannot figure out what to do. My professor provided the final answer (-466kJ), but I need help getting to that answer. Thanks. - 3857.니니43•20moles5. 1.02 g of Mg was reacted with excess 1 M HCI(ag) (255.0 g) in a coffee-cup calorimeter. Tsoln rose from 22.0 °C to41.6°C. Calculate the AHpn of this reaction. GHCI = 3.90 J g-1°C-1 ans: -466 kJ/molLA 8.582750lution=mCATIW Solict or tol rixFa onimmeteb of woled nevig selgisritne notoser bisbnsta erd saUneviL8.0es+= XY HALX 8.Ter-anxT HAibeto 1ol how uoy wods yheelo ieusucY6. A 32.5 g piece of aluminum (which has a molar heat capacity of 24.03 J/(°C-mol)) is heated to 82.4°C and droppedinto a calorimeter containing water (specific heat capacity of water is 4.18 J/(g°C)) initially at 22.3°C. The finaltemperature of the mixture is 24.2°C. Calculate the mass of water in the calorimeter. Ans: 212 gmoini lulseU0.88S-um uoYttonmolo

We know , q(HCl) = m×c×(t2-t1) where t2 = final temperature…

Answered: 5. 1.02 g of Mg was reacted with excess…

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The question I need help with is #5 on the attached image. How do I calculate the change in heat using the given information in #5? I have tried multiple times and cannot figure out what to do. My professor provided the final answer (-466kJ), but I need help getting to that answer. Thanks.

- 3857.니니43
•20moles
5. 1.02 g of Mg was reacted with excess 1 M HCI(ag) (255.0 g) in a coffee-cup calorimeter. Tsoln rose from 22.0 °C to
41.6°C. Calculate the AHpn of this reaction. GHCI = 3.90 J g-1°C-1 ans: -466 kJ/mol
LA 8.582750lution=mCATIW S
olict or tol rixFa onimmeteb of woled nevig selgisritne notoser bisbnsta erd saU
nevi
L8.0es+= XY HA
LX 8.Ter-anxT HA
ibeto 1ol how uoy wods yheelo ieusucY
6. A 32.5 g piece of aluminum (which has a molar heat capacity of 24.03 J/(°C-mol)) is heated to 82.4°C and dropped
into a calorimeter containing water (specific heat capacity of water is 4.18 J/(g°C)) initially at 22.3°C. The final
temperature of the mixture is 24.2°C. Calculate the mass of water in the calorimeter. Ans: 212 g
moini lulseU
0.88S-
um uoY
tton
molo

Expert Solution

Step 1

We know ,

q(HCl) = m×c×(t₂-t₁) where t₂= final temperature

Given m = 255 gm t₁= initail temperature

c = 3.90 j.g^-1c ^-1

t₂= 41.6 c

t₁= 22.0 c

Now

q(HCl) = m×s×(t2-t1)

= 255 gm × 3.90 j.gm^-1 c^-1×(41.6-22.0) c

= 19492.2 j

q_(reaction) = - q(HCl) = -19492.2 j

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