copper compounds are used in large amounts to control the growth of algae and other aquatic plants. copper interferes with photosynthesis in these organisms. what would be the most likely result of an accidental spill of these compounds in an aquatic environment?

Answers

Answer 1

The most likely result of an accidental spill of copper compounds in an aquatic environment is that the growth of algae and other aquatic plants would be controlled by copper.

Copper interferes with photosynthesis in these organisms, which means that it prevents them from producing energy from the sunlight. This can lead to a reduction in the amount of oxygen produced by aquatic plants, which can affect the overall health of the ecosystem.

Copper is toxic to aquatic organisms, so it can cause harm to fish, invertebrates, and other aquatic life forms that are exposed to it. It can also accumulate in sediments and other organic matter, where it can persist for long periods of time. This can lead to the bioaccumulation of copper in the food chain, which can cause further harm to higher trophic levels.

Overall, an accidental spill of copper compounds in an aquatic environment can have significant impacts on the ecosystem and can take a long time to recover from.

Therefore, it is important to use copper compounds responsibly and carefully and to take appropriate measures to prevent accidental spills and leaks.

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Related Questions

A particular tank of oxygen gas contains 785 L at 21°C.If the pressure remains constant,what volume will the gas occupy if the temperature is changed to 28°C? (Write procedure)

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Make sure to list known values on the side. Based on the wording of the question, we know that pressure is constant and moles is constant. You can rearrange the ideal gas equation and solve for the volume.

A particular tank of oxygen gas contains 785 L at 21C.If the pressure remains constant,what volume will

how many moles of carbon dioxide will form when 2.60 moles of pentane react with excess oxygen?

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When 2.60 moles of pentane react with excess oxygen, 13.0 moles of carbon dioxide will form.

The balanced equation for the combustion of pentane (C₅H₁₂) with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O) is:

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

From the balanced equation, we can see that for every 1 mole of pentane, 5 moles of carbon dioxide are produced. Therefore, if 2.60 moles of pentane react, we can calculate the moles of carbon dioxide formed using the mole ratio:

2.60 moles C₅H₁₂ x (5 moles CO₂ / 1 mole C₅H₁₂) = 13.0 moles CO₂

Therefore, when 2.60 moles of pentane react with excess oxygen, 13.0 moles of carbon dioxide will form.

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choose the following true statement. group of answer choices a triple bond may consist of one sigma bond and two pi bonds or of two sigma bonds and one pi bond. a carbon atom involved in only single bonds may not be sp2 hybridized. a pi bond can hold 4 electrons, two above and two below the sigma-bond axis. a sigma bond is a bond resulting from side-to-side overlap of atomic orbitals. a pi bond is a bond resulting from side-to-side overlap of atomic orbitals.

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A triple bond may consist of one sigma bond and two pi bonds.

Sigma bonds are the most powerful kind of covalent chemical bond. they're fashioned by way of head-on overlapping among atomic orbitals. Sigma bonding is most surely described for diatomic molecules the usage of the language and equipment of symmetry businesses

Sigma bond is a covalent bond formed by overlap of atomic orbitals and hybrid orbitals along the bond axis. The sigma bond in the a hydrogen molecule is formed by overlap of a pair of 1s orbitals, one from each hydrogen atom.

The pi bond is the same as that of the p orbital when seen down the bond axis.

A pi bond is a weaker chemical covalent bond than a sigma bond.

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Fill In the
476 nm = 4.76 x 10^? Cm

Recall that 1 nm 1 x 10 m and 1 cm = 1 x 10-2 m. You will want to memorize the metric conversion factors and those covered in this lesson!

Answers

Answer:

I gotthe answer

Explanation:

The explanation is This

Now writing the value in sentific term

It would look like

$4.76 \times {10}^{2} \times {10}^{ - 9} $

$4.76 \times {10}^{2 - 9} $

4.76

$4.76 \times {10}^{ - 7} $

Now in centimeter

$4.76 \times {10}^{ - 5} $

PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE

The Liquified Petroleum Gas (LPG) has the composition of 60% Propane (C 3
H 8
) and 40% Butane (C 4
H 10
) by volume: (a) Find the wet volumetric and gravimetric analysis of the products of combustion when the equivalence ratio (Φ)=1.0. (b) What is the stoichiometric air to fuel ratio for the LPG.

Answers

(a) To find the wet volumetric and gravimetric analysis of the products of combustion when the equivalence ratio (Φ) is 1.0, we need to consider the stoichiometry of the combustion reaction for propane (C₃H₈) and butane (C₄H₁₀).

The balanced combustion reaction for propane can be represented as:
C₃H₈ + (5/2)O₂ → 3CO₂ + 4H₂O

And the balanced combustion reaction for butane can be represented as:
C₄H₁₀ + (6.5)O₂ → 4CO₂ + 5H₂O

Since LPG is composed of 60% propane and 40% butane by volume, we can calculate the wet volumetric and gravimetric analysis based on these proportions.

Wet volumetric analysis:
For the wet volumetric analysis, we consider the volume of the products of combustion relative to the volume of the LPG consumed.

Propane (C₃H₈):
The stoichiometric coefficient of propane in the combustion reaction is 3. Therefore, for every mole of propane burned, we will have 3 moles of CO₂ and 4 moles of H₂O formed.

Butane (C₄H₁₀):
The stoichiometric coefficient of butane in the combustion reaction is 4. Therefore, for every mole of butane burned, we will have 4 moles of CO₂ and 5 moles of H₂O formed.

Considering the initial composition of 60% propane and 40% butane by volume, we can calculate the volumetric composition of the products of combustion:

Volumetric composition of CO₂:
(0.6 * 3) + (0.4 * 4) = 3.6

Volumetric composition of H₂O:
(0.6 * 4) + (0.4 * 5) = 4.6

Therefore, the wet volumetric analysis of the products of combustion is 3.6 parts CO₂ to 4.6 parts H₂O.

Wet gravimetric analysis:
For the wet gravimetric analysis, we consider the mass of the products of combustion relative to the mass of the LPG consumed.

Using the molar masses of the compounds involved in the combustion reaction:
Molar mass of CO₂ = 44 g/mol
Molar mass of H₂O = 18 g/mol

Gravimetric composition of CO₂:
(0.6 * 3 * 44 g/mol) + (0.4 * 4 * 44 g/mol) = 158.4 g

Gravimetric composition of H₂O:
(0.6 * 4 * 18 g/mol) + (0.4 * 5 * 18 g/mol) = 74.4 g

Therefore, the wet gravimetric analysis of the products of combustion is 158.4 grams CO₂ to 74.4 grams H₂O.

(b) The stoichiometric air to fuel ratio for LPG can be determined based on the balanced combustion equations for propane and butane.

For propane (C₃H₈):
C₃H₈ + (5/2)O₂ → 3CO₂ + 4H₂O

The stoichiometric coefficient for propane is 1, which means we need 5/2 moles of O₂ for every mole of propane.

For butane (C₄H₁₀):
C₄H₁₀ + (6.5)O₂ → 4CO₂ + 5H₂O

Why do metals have a positive charge?
Because they lose protons.
Because they gain protons.
Because they gain electrons.
Because they lose electrons.

Answers

Answer:

D. Because they lose electrons

Explanation:

In neutral state the numbers of proton (+) = numbers of electrons (-)

if there are less electrons, there will be more protons than electrons => positive charge

if there are more electrons, there will be more electrons than protons => negative charge

Please answer the question accordingly :)