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3 years ago

What effect do calcium ions have on the water in ecosystems?

Chemistry

2 answers:

Marysya12 [62]3 years ago

6 0

It's because they add to its hardness . Calcium is naturally dissolved in water

BlackZzzverrR [31]3 years ago

4 0

Answer:

They add to its hardness.

Explanation:

Hello,

Water in ecosystems are likely to have dissolved metallic cations such those alkalies and alkaline earth metals form when present in aqueous systems, in this manner, more specifically speaking, calcium ions promote the formation of unsoluble salts such as calcium carbonate, which lead to the increase of the water's hardness that accounts for such cation's content into the water. Therefore, among the following options:

A. They reduce its pH.

B. They decrease its salinity.

C. They add to its hardness.

D. They increase its acidity.

The correct is C. They add to its hardness.

Best regards.

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Which is the largest group of mollusks?

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Answer:

Gastropoda.

Explanation:

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?

A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol.

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

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3 years ago

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Answer:

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2 years ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

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3 years ago

Atoms are the smallest particles of a compound that still retain the properties of that

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Answer:

the answer is true

Explanation:

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