When a solution of Pb(NO3)2(aq) is mixed with a solution of KI(aq), a precipitate of PbI₂ will form; K⁺ and NO₃⁻ are spectator ions.
<u>Explanation:</u>
When an aqueous solution of lead nitrate (Pb(NO₃)₂ is mixed with aqueous solution of potassium iodide (KI), then there is a precipitate formation of lead iodide (PbI₂), and the potassium (K⁺) ion and nitrate (NO₃⁻) ion acts as spectator ions that is ions do not involved in the reaction.
The reaction can be represented as,
Pb(NO₃)₂(aq) + 2 KI (aq) → PbI₂(s) + 2KNO₃(aq)
The ionic equation can be written as,
Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
On both sides of the equation, we have K⁺ and NO₃⁻ ions, which gets cancelled, and these 2 ions are called as spectator ions.
40 grams ÷ 40.08 grams/moles = 1 mole
Answer:
B. 3.0 g/ml
Explanation:
density formula: mass/volume
15/5=3
Answer:
Explanation:
Graham’s Law applies to the effusion of gases:
The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).
If you have two gases, the ratio of their rates of effusion is
The time for diffusion is inversely proportional to the rate.
Let CO₂ be Gas 1 and O₂ be Gas 2
Data:
M₁ = 44.01
M₂ = 32.00
Calculation
Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
