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mojhsa [17]
3 years ago
9

Which process takes place when large rocks crumble due to oxidation?

Chemistry
2 answers:
jok3333 [9.3K]3 years ago
4 0
Abrasion is the answer
VashaNatasha [74]3 years ago
3 0

Answer:

This question is incomplete but the completed question is below

Which process takes place when large rocks crumble due to oxidation?

A) Erosion

B) Deposition

C) physical weathering

D) chemical weathering​

The correct option is D

Explanation:

Weathering can simply be defined as the breakdown of the top surface of rocks. <u>From this definition, it can be deduced that the question describes a type of weathering</u>. Weathering can be of two types; Physical weathering and chemical weathering.

Physical weathering is caused by physical components of the environment such as temperature, water and some activities of living organisms. Physical weathering does not alter the chemical composition of the rock. A <u>chemical weathering</u> is caused by a reaction between the components of the rock and another substance. This type of weathering <u>involves a chemical reaction (for example oxidation)</u> which leads to the breakdown of the rock.

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Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a
kolezko [41]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= 25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x \frac{40.67 KJ}{mol H20} x \frac{1 mol H20}{18 g}

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

3 0
4 years ago
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