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kifflom [539]
3 years ago
14

Distinguish air mass and a front

Chemistry
1 answer:
Readme [11.4K]3 years ago
3 0
Air mass is how much matter is in air. Front is the side that something faces forward with.
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Polar air masses bring warm temperature and have lower air temperature? T or F
snow_tiger [21]

Answer:

t? im pretty sure have a good day

3 0
3 years ago
Study the image and answer the question.
konstantin123 [22]
The image represents A COMPOUND because the molecules are BONDED CHEMICALLY.
A compound is a substance formed when two or more elements combine together chemically. In the process of chemical combination, the chemical bonds that were present in the participating elements will be broken and new chemical bonds will be formed in the product.
7 0
3 years ago
It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichlorometha
Allushta [10]

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

CH_4+CCl_4\rightarrow 2CH_2Cl_2

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = (\frac{actual}{theoretical})100

65.5=(\frac{2.25}{theoretical})100

theoretical=(\frac{2.25(100)}{65.5})

theoretical = 3.44 g

From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.

Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

3.44gCH_2Cl_2(\frac{1molCH_2Cl_2}{84.93gCH_2Cl_2})(\frac{1molCCl_4}{2molCH_2Cl_2})(\frac{153.82gCCl_4}{1molCCl_4})

= 3.12gCCl_4

So, 3.12 grams of carbon tetrachloride are needed to be reacted.

8 0
3 years ago
Can some one help me with this soon as possible please will give brainliest
lana66690 [7]

1. C

2. C

3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit. 

4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust.  Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress. 


I hope this helped :D

7 0
3 years ago
How do you solve this ?NH4+NO4->N2+O2+H2O2
ololo11 [35]
I'm assuming you wanted this equation balanced? If you typed it correctly, it's already balanced for you. :)
8 0
3 years ago
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