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Papessa [141]
3 years ago
14

Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when di

ssolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs. Use the results of your in-lab observations, as well as your textbook and the provided Supplemental Information on pages 65-66, to support your conclusions. NaCI Na2SO4 NaHCO K3PO4 sodium citrate CaCl2 2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na:C204, is the sodium salt of this acid. Categorize it as acidic, basic, or neutral i water. Does the salt dissolve, dissociate, or both in aqueous solutions? What about oxalic acid; does it dissolve, dissociate, or both in water? Explain.
Chemistry
1 answer:
Gwar [14]3 years ago
4 0

Answer:

NaCl neutral

Na₂SO₄ basic

NaHCO₃ basic

K₃PO₄ basic

Na₃C₆H₅O₇ basic

CaCl₂ neutral

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water.

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water

Explanation:

<em>Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs.</em>

<em />

NaCl

When NaCl is dissolved in water the resulting solution is neutral.

Na₂SO₄

When Na₂SO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the SO₄²⁻ ion.

SO₄²⁻(aq) + H₂O(l) ⇄ HSO₄⁻(aq) + OH⁻(aq)

NaHCO₃

When NaHCO₃ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the HCO₃⁻ ion.

HCO₃⁻(aq) + H₂O(l) ⇄ H₂CO₃(aq) + OH⁻(aq)

K₃PO₄

When K₃PO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the PO₄³⁻ ion.

PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

Na₃C₆H₅O₇ (sodium citrate)

When Na₃C₆H₅O₇ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the C₆H₅O₇³⁻ ion.

C₆H₅O₇³⁻(aq) + H₂O(l) ⇄ C₆H₆O₇²⁻(aq) + OH⁻(aq)

CaCl₂

When CaCl₂ is dissolved in water the resulting solution is neutral.

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water according to the following equation:

Na₂C₂O₄(aq) ⇄ 2 Na⁺(aq) + C₂O₄²⁻(aq)

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water according to the following equation:

C₂H₂O₄(aq) ⇄ C₂HO₄⁻(aq) + H⁺(aq)

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What is the mass of 2.40 moles of magnesium chloride, mgcl2
LiRa [457]

Hey there!

MgCl₂

Find molar mass of magnesium chloride.

Mg: 1 x 24.305

Cl: 2 x 35.453

--------------------  

          95.211 grams

One mole of magnesium chloride has a mass of 95.211 grams.

We have 2.40 moles.

2.40 x 95.211 = 228.5

To 3 sig figs this is 229.

The mass of 2.40 moles of magnesium chloride is 229 grams.

Hope this helps!

4 0
3 years ago
Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.
melomori [17]

Answer:

0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, \frac{25.0g}{1} × \frac{1 mol}{56.106g} = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= \frac{0.4558mol}{3.00L} = 0.1485M ≈ 0.15M

5 0
2 years ago
What is the mass of 4.75 mol H2SO4
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M(H2SO4)=n*M=4.75*98=465.5g
7 0
3 years ago
Calculate the freezing point of the solution containing 0.105 m k2s. calculate the boiling point of the solution above. -g
Licemer1 [7]
Quantity of K2S m = 0.105 m 
Number of ions i = 2(K) + 1(S) = 3
 Freezing point depression constant of water Kf = 1.86 
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
 Freezing point = 0 - 0.586 = 0.586 C
 Boiling point constant of water Kb = 0.512
 delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
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3 years ago
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