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3 years ago

Lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI net ionic equation:

Chemistry

1 answer:

MissTica3 years ago

7 0

Answer:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

General Formulas and Concepts:

Solubility Rules
Reaction Prediction

Explanation:

Step 1: RxN

Pb(NO₃)₂ (aq) + KI (aq) → PbI₂ (s) + KNO₃ (aq)

Step 2: Balance RxN

Pb(NO₃)₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq)

Step 3: Ionic Equations

Total Ionic Equation:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq)

Cancel out spectator ions.

Net Ionic Equation:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

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A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?

igor_vitrenko [27]

Answer:

1.15 M

Explanation:

Step 1: Given data

Initial volume (V₁): 0.125 L

Initial concentration (C₁): 3.00 M

Final volume (V₂): 0.325 L

Final concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

8 0

3 years ago

How many moles are present in 25.0 grams of potassium permanganate, KMnO,?

poizon [28]

Answer: 0.158

Explanation:

7 0

3 years ago

When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3

IRISSAK [1]

Answer: The molecular mass of this compound is 131 g/mol

Explanation:

Depression in freezing point:

$\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$\Delta T_f$ = depression in freezing point = $3.9^oC$

$k_f$ = freezing point constant = $20.8^0C/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 0.49 g

$w_1$ = mass of solvent (cyclohexane) = 20.00 g

$M_2$ = molar mass of solute = ?

Now put all the given values in the above formula, we get:

$(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}$

$M_2=131g/mol$

Therefore, the molar mass of solute is 131 g/mol

7 0

3 years ago

Why does every sample of water have the same boiling point?

Tamiku [17]

Answer:

beacuse water is same every where because it is the combination of h2 +o2 h2o which doesn't change while it is different

Explanation:

8 0

3 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago