Gimme ur bath water plzzzz

In a redox reaction, which ion would you expect to be more easily reduced? In a redox reaction, which ion would you expect to be

Levart [38]

Answer:

In a redox reaction, hypochlorite ion would be more easily reduced.

Explanation:

Hypochlorite compounds are generally unstable - e.g. sodium hypochlorite is not available in solid form, because removing water from a NaClO disolution, causes a dismutation, turning it in a mixture of NaCl y NaClO3. Heating the same disolutions will also cause this effect. Hypochlorite will decompose under solar light into chlorides and oxygen.

Due to their low stability, hypochlorite compounds (i.e. those containing the hypochlorite ion) are very strong oxidants. Perchlorate ion can balance its negative charge instead and thus will be less prone to reduction than hypochlorite.

7 0

3 years ago

Which procedure confirms the presence of F ions, CI ions, Br ions, I ions?

nexus9112 [7]

Answer:

F-, no precipitate. Cl-, white precipitate. Br-, very pale cream precipitate. I-, very pale yellow precipitate.

Explanation:

5 0

3 years ago

For the following reaction, 5.61 grams of carbon monoxide are mixed with excess water . Assume that the percent yield of carbon

stealth61 [152]

Answer: The ideal yield of carbon dioxide is 7.506 grams

Explanation:

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

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

Given mass of carbon monoxide = 5.61 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol$

The chemical equation for the reaction of carbon monoxide and water follows:

$CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of carbon monoxide produces 1 mole of carbon dioxide

So, 0.200 moles of carbon monoxide will produce = $\frac{1}{1}\times 0.200=0.200mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.200 moles

Putting values in equation 1, we get:

$0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g$

To calculate the experimental yield of carbon dioxide, we use the equation:

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

Percentage yield of carbon dioxide = 85.3 %

Theoretical yield of carbon dioxide = 8.8 g

Putting values in above equation, we get:

$85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g$