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Paul [167]
1 year ago
8

A local plant nursery uses large sprinklers to water the plants twice a day. the water contains phosphorus, which is a fertilize

r that helps plants grow. some of the water becomes runoff and ends up in nearby streams and lakes. this is an example of which type of short-term human-induced environmental change? eutrophication non-sustainable harvesting nonnative species introduction deforestation
Chemistry
1 answer:
Mumz [18]1 year ago
3 0

Eutrophication

In an aged aquatic habitat like a lake, eutrophication is the progressive rise in the concentration of phosphorus, nitrogen, and other plant nutrients. As the volume of organic matter that can be converted into nutrients increases, the productivity or fertility of such an ecosystem also naturally rises.

<h3>What is Eutrophication ?</h3>

Eutrophication may be caused by a number of things, including overuse of fertilisers, untreated sewage, the use of phosphorous-containing detergents, and industrial waste discharge.

  • Eutrophication naturally. Natural eutrophication is a process that develops in water resources over a very long period of time as a result of a slow buildup of nutrients and organic waste. Anthropogenic or cultural eutrophication.

Learn more about Eutrophication here:

brainly.com/question/26956972

#SPJ4

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A child pushes a desk with a force of 15 N to the right. The desk accelerates to the right. Which of the following statements co
ratelena [41]

Answer:

The answer to your question is: number 1

Explanation:

Third law of Newton: says that for every action ,there is an equal and opposite reaction.

So, if the child is pushing an object to the right, it will recipe the same amount of force that he is exerting to the object but in opposite direction.

Number 2 and 3 are incorrect because, because the third law of Newton says "an equal and opposite reaction", not slightly more or less.

Number 4 is wrong, it is not in agreement with Newton's third law of motion.

5 0
3 years ago
Classify the following as homogeneous or heterogeneous mixtures, or pure
victus00 [196]

Explanation:

Sugar -  Pure substance

Magnesium Ribbon - Pure Substance

Vegetable soup  Heterogeneous mixture

Bath oil - Homogeneous mixture

Tin of assorted biscuits - Heterogeneous mixture

Peanuts and raisins - Heterogeneous mixture

Copper wire - Pure Substance

Bicarbonate of soda  (Baking soda)​ - Pure Substance

6 0
2 years ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
kaheart [24]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

6 0
2 years ago
A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and thi
dezoksy [38]

Answer:

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba(NO3)2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq)

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba(NO3)2 = 0.0148 L * 0.860 M

Moles Ba(NO3)2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba(NO3)2 is the limiting reactant. It will completely be consumed. (0.012728 moles) . K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba(NO3)2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 =  0.012728 moles *  233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield ) * 100 %

% yield = ( 2.57 grams / 2.97 grams ) * 100 %

% yield = 86.5 %

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

7 0
3 years ago
Hello how are u guys today?
Lyrx [107]

Answer:

hello

Explanation:

I am good and you , hope your doing great lol

8 0
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