Answer:
It’s twice as much as 9+10
Explanation:
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
Answer:
add 7.5L of water
Explanation:
M1×V1=M2×V2
M is molarity, V is volume
0.7 × 10 = 0.4 × V2
V2= 17.5L
vol. of water to add= 17.5 - 10 = 7.5L
