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Papessa [141]
2 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]2 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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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
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Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

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            Kc = 1.54e - 31 / 2.61e - 24

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3 years ago
What is a substance?
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7 0
3 years ago
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3 0
2 years ago
Determine the temperature change when 150. G block of gold is supplied with 1.00 • 10 ^3 J of heat
IceJOKER [234]

Answer:

∇T = 51.68°C

Explanation:

Mass = 150g

Heat Energy (Q) = 1.0*10³J

Change in temperature ∇T = ?

Q = mc∇T

Q = heat energy

M = mass

C = specific heat capacity of the gold = 0.129j/g°C

∇T = change in temperature

Q = Mc∇T

1.0*10³ = 150 * 0.129 * ∇T

1000 = 19.35∇T

Solve for ∇T

∇T = 1000 / 19.35

∇T = 51.679°C = 51.68°C

The change in temperature of gold was 51.68°C

8 0
3 years ago
A 10.-liter flask at a given temperature and pressure contains 6.0 × 1023 molecules of hydrogen gas. Under the same conditions o
valentina_108 [34]

Answer:

The same number of molecules, 6.0 × 10²³ molecules.

Explanation:

The amount of any given gas that can be stored in a container depends on the <u>temperature, pressure and volume </u>of the container. It does not depend on the nature (or identity) of the gas.

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