Atoms are the basic building blocks of ordinary matter. Atoms can join together to form molecules, which in turn form most of the objects around you.
Equal. Good luck with other questions!
D.) It depends cuz no yeild is 100%..I mean side reactions also occur in most of the reactions. So mass of the reactant is not equal to the mass of the product. Hope it helps
Answer:
The answer will be 2.98K
Explanation:
Using the formula:
Q = mc∆T
Q= 5,800 (heat in joules)
m= convert 15.2kg to g which is 15200g (mass in grams)
c= 0.128 J/g °c (Specific heat capacity)
∆T= what we need to find (temperature change)
5800J = 15200g x 0.128 x ∆T
= 2.98K
Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
![Q = c \times m \times \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20c%20%5Ctimes%20m%20%5Ctimes%20%5CDelta%20T)
where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
![Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ](https://tex.z-dn.net/?f=Q%20%3D%204.184J%2Fg.%5C%C2%B0C%20%5Ctimes%2075.0g%20%5Ctimes%20%2895.00%5C%C2%B0C%20-%2025.00%5C%C2%B0C%29%20%3D%202.20%20%5Ctimes%2010%5E%7B3%7D%20J%20%3D%202.20%20kJ)
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
![\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol](https://tex.z-dn.net/?f=%5Cfrac%7B-22.0kJ%7D%7B64.00g%7D%20%5Ctimes%20%5Cfrac%7B16.04g%7D%7Bmol%7D%20%3D%20-5.51%20kJ%2Fmol)