Answer:
The correct option is: AgNO₃(aq) + KCl(aq) = AgCl(s) + KNO₃(aq)
Explanation:
Precipitation reaction is a chemical reaction that involves reaction between <em>two soluble salts to give an insoluble salt.</em> This <u>insoluble salt exists as a solid</u> and settles down.
Therefore, the solid formed in a precipitation reaction is known as the precipitate.
As the solid silver nitrate (AgNO₃) and solid potassium chloride (KCl) are <u>soluble in water</u>, therefore, their aqueous solutions are represented as AgNO₃(aq) and KCl(aq), respectively.
The precipitation reaction of AgNO₃(aq) and KCl(aq) gives an <u>insoluble salt, silver chloride (AgCl) and a soluble salt, potassium nitrate (KNO₃).</u>
The insoluble salt, <u>AgCl is called the precipitate</u> and is represented as AgCl(s). Whereas, the <u>soluble salt</u>, KNO₃ is represented as KNO₃ (aq).
<u>Therefore, the chemical equation for this precipitation reaction is:</u>
AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq)
About 8.0 moles of methane.Number of moles = MassMolar mass.
And thus we get the quotient:
128.3⋅g16.04⋅g⋅mol−1=8.0⋅moles of methane.
Note that the expression is dimensionally consistent, we wanted an answer in moles, and the quotients gives, 1mol−1=11mol=mol as required.
Answer:
I think you meant metabolism
Explanation:
me·tab·o·lism
the chemical processes that occur within a living organism in order to maintain life.
168.96 g of carbon dioxide (CO₂)
Explanation:
The chemical reaction representing the combustion of acetylene:
2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)
number of moles = mass / molecular weight
number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)
then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)
X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)
mass = number of moles × molecular weight
mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g
Learn more about:
combustion of hydrocarbons
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