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3 years ago

Write a balanced half-reaction describing the oxidation of solid chromium to aqueous chromium(IV) cations.

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:

Cr(s) ⟶ Cr⁴⁺(aq) + 4e⁻

Explanation:

1. Write the skeleton half-reaction

Cr(s) ⟶ Cr⁴⁺(aq)

2. Balance charge

Add electrons to the side that needs them.

You have 4+ on the right and 0 on the left. You must add 4e⁻ to the right to balance the charge.

Cr(s) ⟶ Cr⁴⁺(aq) + 4e⁻

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How does burning affect an object's property? Burning can cause an object to increase in volume. Burning changes the chemical ma

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Answer: Burning changes the chemical make up of an object.

Explanation:

A chemical change can be defined as a change in the substance when it combines with other kind of substance to form a new substance. A chemical change can also occur when a substance is broken down into two or more products. These changes cannot be reversed. These changes affect the physical make up of an object. For example, burning as when an object is burned it cannot be transformed into its original form. A wood if burned can be converted into ash, water and carbon dioxide cannot regain its original form after burning so burning brings about chemical change in an object.

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3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

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Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

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