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mixas84 [53]
3 years ago
5

Which statement best compares coastal ecosystems to open ocean ecosystems?

Chemistry
2 answers:
nasty-shy [4]3 years ago
7 0

The answer is: Coastal ecosystems have more sunlight, more nutrients, and higher levels of productivity than open ocean ecosystems.

Coastal ecosystems are areas where land and water join (bays, estuaries, salt marshes, mangroves, wetlands).

In coastal ecosystems live many different types of plants and animals.

The ocean is divided into four major zones: the intertidal, neritic, oceanic and abyssal, each zone includes several ecosystems, which have adapted to specific habitats.

Anit [1.1K]3 years ago
4 0
Coastal ecosystems have more sunlight, more nutrients, and higher levels of productivity than open ocean ecosystems. These areas are shallow (lots of light) and close to river mouths (bringing high amounts of nutrients)
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A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c
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3 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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