Answer:
Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)
Explanation:
Let's consider the following molecular equation.
Mg(s) + SnSO₄(aq) ⇄ MgSO₄(aq) + Sn(s)
The full ionic equation includes al the ions and the species that do not dissociate in water.
Mg(s) + Sn²⁺(aq) + SO₄²⁻(aq) ⇄ Mg²⁺(aq) + SO₄²⁻(aq) + Sn(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the species that do not dissociate in water.
Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)
The balanced equation is:
BaCl2 (aq) + Na2SO4 (aq) ----> BaSO4(s)+ 2 NaCl(aq)
This is a double replacement reaction.
The reactants are:
a) BaCl2: barium chloride, a ionic compound, therefore soluble in water,
b) Na2SO4: sodium sulfate, another ionic compound, therefore also soluble in water.
The products are:
c) BaSO4: barium sulfate, a solid not soluble in water which precipitates.
d) NaCl: sodium chloride, an ionic compound, therefore soluble in water.
Sulfuric acid reacts violently with alcohol and water to release heat. It reacts with most metals, particularly when diluted with water, to form flammable hydrogen gas, which may create an explosion hazard. ... Hazardous decomposition products are as follows: sulfur dioxide, sulfur trioxide, and sulfuric acid fumes.
Answer:
A molecule of sucrose (C12H22O11) has 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Explanation:
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<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in = 
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
