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riadik2000 [5.3K]
4 years ago
7

How much solute (in mL) would you need to add to solvent in order to create 825 mL of a final solution that is 21% solute by vol

ume? (round to the nearest whole number and do not include units) *
Chemistry
1 answer:
olchik [2.2K]4 years ago
3 0

Answer:21÷100×825 = 173.25

173

Explanation:

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Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
BARSIC [14]
<h3>Answer:</h3>

              6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

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