Answer:
<h2>
<u>OH-</u> es el símbolo del agua ozonizada.</h2>
Answer:
2.47 M
Explanation:
Molarity is (mol of solute)/(L of solvent). To find molarity, convert grams of NaCl to moles of NaCl and milliliters of water to liters of water.
72.3 g ÷ 58.44 g/mol = 1.2372... mol
500 mL ÷ 1000 mL = 0.5 L
Molarity = 1.2372/0.5 = 2.47 M
10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
What ois Theoretical yield ?
The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.
The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.
- As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃
Therefore ,
10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂
- As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃
Therefore ,
10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃
Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
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The number of hours required : 37.2 hours
<h3>Further explanation</h3>
Given
⁴²K (potassium -42)
Required
The number of hours
Solution
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours
For three half-life periods :
In electrophilic aromatic substitution reactions the hydroxyl group is an o,p-director because: hydroxyl group donates the electron density to the ring by induction and destabilizes the meta sigma complex and by resonance and it stabilizes the ortho and para sigma complexes of aromatic ring .
Most ring activators have atoms with unshared electron pairs directly attached to a carbon atom of the benzene ring . For example, the — OH group has two pairs of unshared electrons on the oxygen atom , which will form a bond to a carbon atom of the benzene ring . Thus , the — OH group will be an activating group in electrophilic aromatic substitution reactions .
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