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2 years ago

Which of these humans activities is a serious threat to marine ecosystems

Chemistry

2 answers:

xxTIMURxx [149]2 years ago

8 0

Answer:

Pollution, non native species being introduced, and other human activities

Explanation:

Human activities affect marine ecosystems as a result of pollution, overfishing, the introduction of invasive species, and acidification, which all impact on the marine food web and may lead to largely unknown consequences for the biodiversity and survival of marine life forms.

Paraphin [41]2 years ago

5 0

Answer:

Pollution, non native species being introduced, and other human activities

Explanation:

Human activities affect marine ecosystems as a result of pollution, overfishing, the introduction of invasive species,and acidification, which all impact on the marine food web and may lead to largely unknown consequences for the biodiversity and survival of marine life forms.

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Write a net ionic equation to show how codeine, c18h21o3n, behaves as a base in water.

Lera25 [3.4K]

Actually, the ionic equation for this is a reversible equation since codeine is a weak base. Any weak base or weak acids do not completely dissociate which makes them a reversible process. The ionic equation for this case is:

<span>C18H21O3N + H3O+ </span><=> C18H21O3NH+ + H2O

7 0

3 years ago

Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.

Naddik [55]

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M

3 0

3 years ago

Which type of internal structure represents a gemstone.

Lemur [1.5K]

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3 0

2 years ago

What is an extinction event

aliya0001 [1]

An extinction event is a widespread and rapid decrease in the biodiversity on Earth. Such an event is identified by a sharp change in the diversity and abundance of multicellular organisms. It occurs when the rate of extinction increases with respect to the rate of speciation. -Wikipedia

4 0

3 years ago

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

3 years ago