I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.
Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
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.
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.
The answer is B, early in the morning water covered areas (lakes,ponds,puddles,etc.) will vaporize a little bit because of the heat from the sun and it will continue all day, vapors rise towards the atmosphere and since it's a lot cooler there it will condense into a cloud which is full of tiny frozen water particles. Hope this helps <span />
A have no potential energy
Answer:
The O is being oxidized, but at the same time, is being reducted.
Explanation:
H₂O₂(l) + ClO₂(aq) → ClO₂(aq) + O₂(g)
In this reaction, we have 4 compounds:
Hydrogen peroxide
Chlorine dioxide (twice)
Oxygen
In both dioxide, the Cl acts with +4 in oxidation state; the oxygen acts with -2.
Oxgen in ground state has 0, as oxidation number.
In peroxide, the H acts with +1 but the oxygen acts with -1.
Peroxide is making the oxidation number from the O in the ClO₂, to decrease (reduction) and to increase in the O, at the ground state.
Hydrogen peroxide is a good reducing and oxidizing agent at the same time.
