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goblinko [34]
3 years ago
14

What happens to an electron during an electron transition?

Chemistry
1 answer:
ICE Princess25 [194]3 years ago
4 0
I think the answer is d not real fir sure I'll find out in the morning sorry
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The three primary sources of air pollution are _____.
Kisachek [45]

There are four main types of air pollution sources:

<span>mobile sources – such as cars, buses, planes, trucks, and trainsstationary sources – such as power plants, oil refineries, industrial facilities, and factoriesarea sources – such as agricultural areas, cities, and wood burning fireplaces<span>natural sources – such as wind-blown dust, wildfires, and volcanoes........hope i helped</span></span>
7 0
3 years ago
Read 2 more answers
Help and show work please
olya-2409 [2.1K]
ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :

•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32

•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius

•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin





4 0
3 years ago
What is the molar mass of sulfur (S)?
Lesechka [4]

Answer:

A. 32.06 g/mol

Explanation:

The molar mass units are always g/mol

8 0
2 years ago
Read 2 more answers
The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} $

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

6 0
2 years ago
The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi
cricket20 [7]

Answer:

See explanation

Explanation:

Number of moles of each gas is

Nitrogen = 15/28 = 0.536 kmoles

Helium = 5/4  = 1.25 kmoles

Methane = 60/16 = 3.75 kmoles

Ethane =    20/30 = 0.67 kmoles

Total number of moles =  0.536 kmoles + 1.25 kmoles + 3.75 kmoles +  0.67 kmoles = 6.206 kmoles

Mole fraction of each gas;

Nitrogen = 0.536 kmoles/6.206 kmoles = 0.086

Helium = 1.25 kmoles/6.206 kmoles = 0.201

Methane = 3.75 kmoles/6.206 kmoles =0.604

Ethane = 0.67 kmoles/6.206 kmoles =0.108

Partial pressure of each gas;

Nitrogen = 0.086 * 1200 kPa = 103.2 kPa

Helium = 0.201 * 1200 kPa = 241.2 kPa

Methane = 0.604 * 1200 kPa = 724.8 kPa

Ethane = 0.108 * 1200 kPa = 129.6 kPa

Apparent specific heat at constant pressure;

Cp = (0.15 * 1.039) + (0.05 * 5.1926) + ( 0.6 * 2.2537) + (0.2 * 1.7662)

Cp = 2.12 KJ Kg-1 K-1

Cv = Cp- Ru/M

Cv= 2.12 - 8.314/16.12 = 1.604 KJ Kg-1 K-1

8 0
2 years ago
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