Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:

= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660](
- 23.32)
(
- 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
I think that you have put up an incomplete question. However, i am answering the question based on my research and knowledge.
Lissa- accuracy and precision are both low
Lamont- accuracy and precision are definitely high
<span>Leigh Anne- accuracy is low but precision is definitely high.
</span>
I hope that this is the answer that you were looking for and the answer has definitely come to your desired help.
Answer:
V₂ = 5.97 L
Explanation:
Given data:
Initial temperature = 9°C (9+273 = 282 K)
Initial volume of gas = 6.17 L
Final volume of gas = ?
Final temperature = standard = 273 K
Solution:
Formula:
The Charles Law will be apply to solve the given problem.
According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 6.17 L × 273K / 282 k
V₂ = 1684.41 L.K / 282 K
V₂ = 5.97 L
I hope you understood
hit me up if you have any other questions :>
Answer:
50.8 g
Explanation:
Equation of reaction.

From the given information, the number of moles of methane = mass/ molar mass
= 15.4 g / 16.04 g/mol
= 0.960 mol
number of moles of oxygen gas = 90.3 g / 32 g/ mol
= 2.82 mol
Since 1 mol of methane requires 2 moles of oxygen
Then 0.960 mol of methane will require = 0.960 mol × 2 = 1.92 mol of oxygen gas
Thus, methane serves as a limiting reagent.
2.82 mol oxygen gas will result in 2.82 moles of water
So, the theoretical yield of water = moles × molar mass
= 2.82 mol × 18.01528 g/mol
= 50.8 g