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3 years ago

List at least three scientific disciplines related to chemistry mentioned or alluded to in the article. The Ozone Hole

Chemistry

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

Answer: biology, climate Science, astronomy etc.

Explanation:

The scientific disciplines that are related to chemistry mentioned or alluded to in the article are:

• Biology: Biology is a natural science which studies living organisms, and this include their molecular interactions, evolution, physical structure, and chemical processes.

• Astronomy: Astronomy is the study of space, and the universe. In astronomy, the stars, planets, and galaxies are all studied.

• Climate science: Climate science is also referred to as climatology and it's the scientific study of climate.

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What will happen when a tree fell in a river

beks73 [17]

It's depending on what size of the tree it is. For example, if the tree is large it will block most of the water that runs through. If the tree is small it shouldn't do as much besides block a little amount of water and it will probably go down stream.
This is based on my thinking

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3 years ago

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3. The same well water contains dissolved calcium (Ca), magnesium (Mg), sodium (Na) and potassium (K) at 5.0, 1.0, 136 and 1.2 m

Verdich [7]

Answer: The concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$ , $4.20\times 10^{-5}mol/L$ , $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively

Explanation:

To convert the mass from milligrams to grams, we use the conversion factor:

1 g = 1000 mg

To convert the given concentration fro grams to moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For Calcium:

Concentration given = 5.0 mg/L = $5.0\times 10^{-3}g/L$

We know that:

Molar mass of calcium element = 40 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{5.0\times 10^{-3}g/L}{40g/mol}\\\\\text{Concentration in mol/L}=1.25\times 10^{-4}mol/L$

For Magnesium:

Concentration given = 1.0 mg/L = $1.0\times 10^{-3}g/L$

We know that:

Molar mass of magnesium element = 24 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.0\times 10^{-3}g/L}{24g/mol}\\\\\text{Concentration in mol/L}=4.20\times 10^{-5}mol/L$

For Sodium:

Concentration given = 136 mg/L = $136\times 10^{-3}g/L$

We know that:

Molar mass of sodium element = 23 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{136\times 10^{-3}g/L}{23g/mol}\\\\\text{Concentration in mol/L}=5.91\times 10^{-3}mol/L$

For Potassium:

Concentration given = 1.2 mg/L = $1.2\times 10^{-3}g/L$

We know that:

Molar mass of potassium element = 39 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.2\times 10^{-3}g/L}{39g/mol}\\\\\text{Concentration in mol/L}=3.08\times 10^{-5}mol/L$

Hence, the concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$ , $4.20\times 10^{-5}mol/L$ , $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively

3 0

3 years ago

In the game of pool, a player uses a stick called a cue to strike a white ball so

o-na [289]

Answer:

the answer is B: cue - white ball - blue ball.

Explanation:

The energy transfers from the stick to the white ball, then from the white ball to the blue ball. I also took this and it was correct!

7 0

3 years ago

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How do halogens differ from noble gases?

Sergeeva-Olga [200]

Halogens differ from Noble Gases because Halogens consist of 5 Elements whereas Noble Gases consist of 18. Also, the elements are different

7 0

3 years ago

A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials an

Vika [28.1K]

Answer:

( $ 8,384 )

Explanation:

- From given information we know that 1 mol mCPHA provides the oxygens to 1 mol of propene, to make 1 mole of C3H6O so:

1 mol C3H6 + 1 mol mCPHA --> 1 mol C3H6O

( Mr = 42.08 g) ( Mr = 172.57 g) ( Mr = 58.08 g )

- For 1 kg equivalent equation by dividing the whole equation by the highest Molar Mass i.e of C3H6O, the result would be:

1 mol C3H6 + 1 mol mCPHA --> 1 mol C3H6O

( 42.08 / 58.08 ) ( 172.57 / 58.08 ) ( 1 )

= ( 0.72452 kg ) ( 2.9712 kg ) ( 1 kg )

- However note that the reaction gives only a 96% yield, we scale up the reactants to get that desired 1 kg of C3H6O:

(0.72452 kg ) (96/100) + (2.9712 kg) (96/100) --> 1 kg

= ( 0.75471 kg ) + ( 3.095 kg ) ---------------> 1 kg

- The costs for each component produced:

(0.75471 kg C3H6) ($10.97 per kg) = $8.279

(3.095 kg mCPHA) ($5.28 per kg) = $16.342

(0.75471 kg C3H6) / (0.0210 kg C3H6 / L dichloromethane) = 35.939

(35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19

- The cost of waste disposal ($5.00 per kilogram of propene oxide) produced total cost, disregarding labor,energy, & facility costs:

$8.279 + $16.342 + $ 76.19 + $5.00 = $105.81 per kg C3H6O produced

- Profit: ($258.25 / kg C3H6O) - ($105.81 cost per kg) = $152.44 profit /kg

- Calculate the profit from producing 55.00kg of propene oxide:

(55.00kg) ($152.44 /kg) = $8,384.2 .. ( $ 8,384 )

6 0

3 years ago