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blsea [12.9K]
3 years ago
13

What are the differences between weathering, erosion, and deposition?

Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0

Answer:

Erosion: the process of eroding or being eroded by wind water or other natural agents.

Weathering: the process of wearing or being worn by long exposure to the atmosphere

Deposition: the action of deposing something

Example: (deposition) " pebbles were formed by the deposition of calcium in the solution"

Explanation:

hope this is right

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Describe one experiment that could be carried out to determine which chemical reaction, decomposition, dimerization, or racemiza
mash [69]

Answer:

Thermal decomposition

Explanation:

The thermal decomposition of amino acids could be carried out at different temperatures and at fixed pressure in a reactor. The effect of temperature on products and pathway could be determined by using time as a function. The hydrothermal decomposition of amino acids takes two key paths

1. The delamination of amino acids to produce ammonia and organi acids

2. The decarboxylation to produce amines and carbonic acids.

The thermal decomposition is and effective experiment that could determine the chemical reaction rate and order of reaction, since it's working with time.

8 0
3 years ago
What amount if force need to be applied to a 20 kg bowling ball to give it an acceleration of 5m/s
balandron [24]

Answer:

100N

Explanation:

Given parameters:

Mass of the bowling ball  = 20kg

Acceleration  = 5m/s²

Unknown:

Amount of force applied  = ?

Solution:

To solve this problem, we apply newton's second law of motion.

    Force  = mass x acceleration

Now insert the parameters and solve;

   Force  = 20 x 5  = 100N

5 0
3 years ago
Will give brainlist need answers asap
Serga [27]

Answer:

40 g

Explanation:

Find the line labeled KClO3 (which might take you a min, theres a lot of lines here)

Notice that when the line creates a direct point, you can measure the exact temperature needed to dissolve a certain amount (like how they gave 30 degrees and it lined up perfectly with the 10 g line. )

Since its asking for the amount at 80 degrees, all you need to do is trace the line to the 80 degree point, and look at the grams. (notice it made a direct point, so there definitely should be any decimals or guesswork)

By reading the graph, you can tell that at 80 degrees, it dissolves 40 grams, and that is your answer.

Hope this helps :)

3 0
3 years ago
In a molecule, how do we show what types of atoms are there and how many atoms there are of each type. ​
Sophie [7]
Chemical or molecular formulas? Symbols from the periodic table are used to show which atoms are present in a molecule and subscripts show the number of atoms and of each type in the molecule.
6 0
3 years ago
1. A mixture of NaOH and Na2CO3 solution required 20.50 mL of 0.5 M HCl using
netineya [11]

Answer:

i) for NaOH (or KOH) = 13.75 mL

ii) for Na_{2}CO_{3} = 13.50 mL.

Explanation:

Phenolphthalein is an indicator which shows change in color when the conditions are highly basic.

Methyl orange is an indicator which shows change in color in presence of highly acidic medium.

For titration of NaOH and HCl we can use phenolphthalein but for sodium carbonate (a weak base) with HCl we use mehtyl orange.

Now in case of mixture of given strong base and weak base the reading or end point obtained from phenolphthalein, shows the neutralization of NaOH only and half neutralization of sodium carbonate.

NaOH+HCl--->NaCl+H_{2}O

Na_{2}CO{3}+HCl--->NaHCO_{3}+H_{2}O

While the reading or end point of methyl orange shows the neutralization of both the base present.

a) The volume of HCl used for phenolphthalein end point = 20.50 mL

Let us say the volume of HCl used for NaOH = V1

The volume of HCl used for half neutralization of sodium carbonate = V2

V1+V2 = 20.50.........(1)

b) the volume of HCl used for methyl organe end point = 27.25

This volume of HCl used for both NaOH and Na₂CO₃

V1+2V2 = 27.25  ... (2)

Equating equation 1 and 2

20.50-V2=27.25-2V2\\V2=6.75

i) Thus the volume of Acid used for NaOH (Or KOH if present in place of NaOH) = 20.50-6.75= 13.75 mL

ii) the volume of acid used for sodium carbonate =2X6.75= 13.5mL

4 0
3 years ago
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