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3 years ago

How did the construction of a parking lot, hiking trail, and playground affect the fish populations in the pond? PLEASE HELP.

Chemistry

2 answers:

Tpy6a [65]3 years ago

4 0

A),b),c),d) answers?

KIM [24]3 years ago

4 0

For construction of parking lot, hiking trail, and playground, area is required which results in the destruction of ponds which further affect the population of fish. Due to destruction of ponds, population of fish decreases.

Construction of parking lot, hiking trail, and playground results in formation of waste and the disposal of these waste in water creates water pollution which is harmful for all aquatic species. Thus, polluted water affect the population of fish badly.

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How many moles of aluminum are needed to react completely with 1.2 mol of feo

Amanda [17]

Answer:

0.8 mol.

Explanation:

The balanced equation for the reaction between Al and FeO is represented as:

2Al + 3FeO → 3Fe + Al₂O₃,

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

Using cross multiplication:

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴ The no. of moles of Al are needed to react completely with 1.2 mol of FeO = (2 mol)(1.2 mol)/(3 mol) = 0.8 mol.

3 0

3 years ago

How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C

rjkz [21]

20600Cal

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

H = m c Ф

c = specific heat capacity of water = 4.186J/g°C

H is the amount of heat

Ф is the change in temperature

H = m c (Ф₂ - Ф₁)

H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

1kilocalorie = 4184J

85996.56J to kCal; $\frac{85996.56}{4184}$ = 20.6kCal = 20600Cal

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0

3 years ago

In the reaction catalyzed by the pyruvate dehydrogenase complex, the two carbons constituting the acetyl group are

lara [203]

Transferred to the lipoamide by an earlier intermediate in the process.

The pyruvate dehydrogenase complex (PDC) is a mitochondrial multienzyme complex composed of three different enzymes

<h3>What reaction is catalyzed by enzyme 2 of the pyruvate dehydrogenase complex ?</h3>

the pyruvate dehydrogenase complex is the bridge between glycolysis and the citric acid cycle

Five coenzymes are used in the pyruvate dehydrogenase complex reactions: thiamine pyrophosphate or TPP, flavin adenine dinucleotide or FAD, coenzyme A or CoA, nicotinamide adenine dinucleotide or NAD, and lipoic acid.
during the reactions catalyzed by the pyruvate dehydrogenase complex, it is first reduced to dihydrolipoamide, a dithiol or the reduced form of the prosthetic group, and then, reoxidized to the cyclic form.

Learn more about Pyruvate dehydrogenase complex here:

brainly.com/question/16346028

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4 0

2 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar