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aksik [14]
3 years ago
15

WARMING OCEAN TEMPERAUTRE AFFECTS THE BIODIVERSITY OF ORGANISMS. WHICH OF THE FOLLOWING DISRUPTIONS WOULD YOU EXPECT TO SEE IN A

POPULATION 0F NORTHERN WHALES AS A RESULT OF WARMER OCEAN WATER?
Chemistry
1 answer:
Ilya [14]3 years ago
7 0

Climate change greatly affects the community of whales and dolphins. Most of the species of whales live in a specific temperature range, because of the non-availability of their food, zoo plankton in warm waters. One of the most important organisms whales prey up on are zoo plankton, whose number declined greatly due to the warming of oceans. These whales migrate to Northern waters overcome the effects of increasing temperature and in search of food. So ,we can expect a disruption in the food chain as a result of the warm oceanic waters.

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According to this balanced equation, how many grams of water (H.0) form in
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your simpal answer is 177.32

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23.9 km + 15 km
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Which statement best expresses Newton's first law of motion? Group of answer choices Objects are constantly speeding up or slowi
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All objects resist changes to their states of motion.  

Explanation:

Sir Issac Newton was an English physicist, mathematician, astronomer and a well known author. Besides, he was a great scientist. Newton discovered many scientific phenomenon and scientific theories in nature.

The most important and famous discoveries of Newton is the Newton's laws of motion. Newtons stated three laws of motion, namely, Newtons 1st law of motion, Newton's 2nd laws of motion and Newton's 3rd law of motion.

According to Newtons's 1st law of motion : A body continues to be in the state of motion or in the state of rest until and unless an external force is applied to it. In other words, all bodies resists changes to the states of their motion or rest.

7 0
3 years ago
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
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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
How to find limiting reagent
Nookie1986 [14]
The reagent which limits the reaction is called limiting reagents.

____

For example:- N2+3H2gives 2NH3.so here nitrogen limits the reaction.

3 0
3 years ago
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