A 10.0 g piece of metal is placed in an insulated calorimeter containing 250.0 g of water initially at 20.0 °C. If the final temperature of the mixture is 25.0 °C, the heat change of water will be 5230 J.
Weight of metal = 10.0 g
Weight of water = 250.0 g
Initial temperature of water (Ti) = 20.0 °C
Final temperature of water (T2) = 25.0 °C
The specific heat capacity of water is 4.184 J/g.°C
We can calculate the heat change of water by the following equation;
<em>Q = mwater × Cwater × (T2-Ti)</em>
Q = 250.0 g × 4.184 J/g.°C × (25.0 - 20.0) °C
Q = 5230 J
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Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
Give people an chance to solve your question. You don’t need to post more than once. Your answer is Fe3
Answer:
Potassium selenide
Explanation:
Potassium selenide (K2Se)