Salt=compound, soda=liquid solution, aluminum foil=element, milk=colloid,
steel=solid solution
An aqueous solution contains the following ions Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ and more than one precipitate will form are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
<h3>What is precipitate?</h3>
Precipitate is the insoluble compound which is present at the bottom of any chemical reaction in the solid state.
If in an aqueous solution Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ ions are present then:
- Compounds AgCl, PbCl₂, PbSO₄ & Ag₂SO₄ are not soluble in water as it is present in the form of precipitate.
- Pb(NO₃)₂ is fully soluble in water and will not make precipitate.
Hence precipitates are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
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The oxidation number sulfur in H₂S is -2.
A compound's total number of oxidations must be zero.
The two hydrogen atoms in the chemical hydrogen sulfide, H₂S, each have an oxidation number of +1, making a total of +2. As a result, the compound's sulfur has an oxidation number of -2, and the total number of oxidations is 0.
Assume that the sulfur atom in H₂S has an oxidation number of x.
S be x.
Now,
2+x=0
⇒x=−2
<h3>What is oxidation number?</h3>
The total number of electrons that an atom either receives or loses in order to create a chemical connection with another atom is known as the oxidation number, also known as the oxidation state.
Depending on whether we are taking into account the electronegativity of the atoms or not, these phrases can occasionally have a distinct meaning. Coordination chemistry commonly makes use of the phrase "oxidation number."
<h3>What distinguishes an oxidation number from an oxidation state?</h3>
In contrast to the oxidation state, which indicates how oxidised an atom is in a molecule, the oxidation number describes the charge that the core metal atom will retain once all ligands have been removed.
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The answer is metal. Metals are always named first in ionic compounds, like KNO3 for example. I hope this helps!
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g