The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
Option (D)
Explanation:
Weathering is usually defined as the disintegration of rocks at the surface of the earth. This break down of rocks is mainly caused by the geological processes that occur on the earth's surface. This process results in the formation of sediments that are transported and deposited in a new environment.
This weathering process primarily takes place in three different ways such as-
- Physical weathering- Here, the rocks are broken down by the physical agents such as wind, water, ice.
- Chemical weathering- Here the rocks are broken down when interacts with the chemical containing water.
- Biological weathering- here, the rocks are broken down due to the activities done by organisms such as plants and animals.
In the given condition, Quincy can see a real example of rock weathering in the high mountainous region, as in the mountainous region the rocks are frequently weathered and eroded by the agents such as wind, water, and ice.
Thus, the correct answer is option (D).
I think it’s d
If it’s wrong then I’m sorry
Melting ice cream is a physical change because it is a phase change, the water in the ice cream gains thermal energy and the inter-molecular forces are overcome and the molecules move further apart forming a liquid. It is not a chemical change because no chemical reactions occurred.