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3 years ago

Which of the following is a chemical change

Chemistry

1 answer:

Oliga [24]3 years ago

3 0

C. a burning candle

all the other choices are physical changes

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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide

Luda [366]

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

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8 0

3 years ago

What type of chemical bond holds the na+ and cl- ions together in table salt?

fredd [130]

<em>The bond between Na + and Cl- ions is an ionic bond</em>

<em>To achieve stability, atoms with low ionization energy such as Na atoms will release electrons and bind to atoms with high-affinity energy such as Cl atoms</em>

$\boxed{\boxed{\bold{Further~explanation}}}$

Atoms have different stability. Unstable atomic atoms will try to form stable electron configurations like those of noble gases. Where noble gases have the number of outer electrons 2 or 8

The formation of electron configurations such as noble gases can be done by forming shared ions or electron pairs

Atoms with high ionization energy and atoms that are difficult to draw electrons will form bonds with shared electron pairs (electrons can be from one atom or two atoms attached)

In forming ions, atoms will release or attract electrons

Ion bonds occur because atoms that have low ionization energy (easily release electrons) form a + ion, and these electrons are bound by atoms that have large affinity energy (easily pulling electrons)

Generally, ionic bonds occur in metal and nonmetallic elements. Metal elements have low ionization energy and non-metallic elements have a high electron affinity

In Na atoms, to achieve stability, Na will release one electron so that it has an electron configuration like noble gas Ne

Na: 2 8 1

Ne: 2 8

so that it becomes a Na + ion

While the Cl atom will bind one electron, so it has an electron configuration like noble gas Ar

Cl: 2 8 7

Ar: 2 8 8

So that it becomes a Cl- ion

Finally, there will be an attractive attraction between positive ions and negative ions and NaCl is formed

So the bond between Na + and Cl- ions is an ionic bond

$\boxed{\boxed{\bold{Learn~more}}}$

the octet rule

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a noble gas

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ionic bonds and covalent bonds

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Keywords: ionic and covalent bonds, the octet rule, noble gases, electron configurations

4 0

3 years ago

Read 2 more answers

Based on its chemical formula, which of the following substances is an organic compound

Tom [10]

I don't understand your question

6 0

3 years ago

Read 2 more answers

What is the purpose of the chemical ammonia (NH3) in hair dyes?

Alinara [238K]

Answer: Ammonia, an alkaline chemical, is used to raise the pH level (it's levels of acidity) of our hair during the colouring process. This then lifts the cuticles of the hair fibre and allows the colour to be deposited onto the cortex (the inner part of the hair protected by the cuticles). In hair coloring products, ammonium hydroxide is used to support the lightening action of hydrogen peroxide and to prepare hair to accept color pigments. Alkaline properties of ammonia raise the cuticle and allow peroxide and dye molecules to penetrate the hair shaft. (creds to the internet)

Explanation: Hope this helps! :D

5 0

3 years ago