Chemistry, 31.12.2019 00:31 angiejtc0908

Need ! calculate the mass of each substance. show your work.
a. 7.00 mil h2o2
b. 5.60 mol naoh
c. o.780 mol ca(cn)2

Answers: 2

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Answers

Answer from: manou76

Explanation:

Given data:

A) Moles of H₂O₂ = 7.00

Mass of H₂O₂ = ?

Solution:

Number of moles = mass/ molar mass

Now we will rearrange this formula:

Mass = Number of moles × molar mass

Mass = 7.00 mol × 34 g/mol

Mass = 238 g

B) Moles of NaOH = 5.60 mol

Mass of NaOH = ?

Solution:

Number of moles = mass/ molar mass

Now we will rearrange this formula:

Mass = Number of moles × molar mass

Mass = 5.60 mol × 40 g/mol

Mass = 224 g

C) Moles of Ca(CN)₂ = 0.780

Mass of Ca(CN)₂ = ?

Solution:

Number of moles = mass/ molar mass

Now we will rearrange this formula:

Mass = Number of moles × molar mass

Mass = 0.780 mol × 92.11 g/mol

Mass = 71.85 g

Answer from: Quest

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.

chemical decomposition is opposite of chemical synthesis.

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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Need ! calculate the mass of each substance. show your work.
a. 7.00 mil h2o2
b. 5.60...

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