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Ira Lisetskai [31]
1 year ago
6

Which is responsible for the formation of rust on metals?(1 point)

Chemistry
1 answer:
Rainbow [258]1 year ago
5 0

The correct answers about rusting, air pollution and products of water with rock are:

Option A. oxidation

Option D. Pollutants mix with air and water to make acid rain.

Option B. clay minerals and calcium carbonate

<h3>What is rusting?</h3>

Rusting is the process by which a metal especially reacts with oxygen in the atmosphere and water vapor to form a hydrated oxide of the metal known as hydrated iron (iii) oxide. This is known as rust.

The process is an oxidation process; option A.

<h3>How does air pollution impact chemical weathering?</h3>

Air pollution is the presence of substances in air known as pollutants which makes the air impure.

Air pollution impact chemical weathering  as the pollutants mix with air and water to make acid rain which weathers rocks; option D.

<h3>What products are produced when water reacts with sodium in rocks?</h3>

The reaction of water with sodium in rocks result in the formation of clay minerals and calcium carbonate also as limestone, marble or chalk; option B.

In conclusion, the presence of pollutants in air results in acid rain and hence rock weathering.

Learn more rock weathering at: brainly.com/question/2341950

#SPJ1

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speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
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In 1860, Chemists could make which of the following statements about the known chemical elements?
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Answer:

b. Some had similar properties

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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
How is 0.00069 written in scientific notation?
vfiekz [6]
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The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo
noname [10]

Answer:

<h2>36.09 L</h2>

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

V_1 =  \frac{P_2V_2}{P_1}  \\

We have

V_1 =  \frac{7.98 \times 14.2}{3.14} =   \frac{113.316}{3.14}  \\  = 36.0878...

We have the final answer as

<h3>36.09 L</h3>

Hope this helps you

8 0
2 years ago
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