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)
Avogadro’s number allows you to change moles to number of molecules/atoms.
Answer:
0.508 L of solution.
Explanation:
Always a safe bet to convert to moles:
Where n is moles, m is mass, and MM is molar mass.
Now remember the equation for concentration (molarity):
Where C is the concentration, n is moles, and V is volume.
To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:
Therefore we can make 0.508 L of solution.
Answer:
The empirical formula of the hydrocarbon is CH.
Explanation:
The following data were obtained from the question:
Mass of hydrocarbon = 2.9 mg
Mass of CO2 = 9.803 mg
Mass of H2O = 2.006 mg
Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.
This is illustrated below:
Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol
Mass of CO2 = 9.803 mg
Mass of C in the compound =?
Mass of C in the compound =
12/44 x 9.803
= 2.674 mg
Molar mass of H2O = (2x1) + 16 = 2 + 16 = 18 g/mol
Mass of H2O = 2.006 mg
Mass of H in the compound =
2/18 x 2.006
= 0.223 mg
Finally, we shall determine the empirical formula of the hydrocarbon as follow:
Carbon (C) = 2.674 mg
Hydrogen (H) = 0.223 mg
Divide by their molar mass
C = 2.674 /12 = 0.223
H = 0.223 / 1 = 0.223
Divide both side by the the smallest
C = 0.223/0.223 = 1
H = 0.223/0.223 = 1
Therefore, the empirical formula of the hydrocarbon is CH.
