<u><em>The answer is definitely the sun</em></u> because a sun is a star. Other stars are to far, so thats why its soo tinny. Some stars are brighter than the sun... But your answer is definitely <u><em>SUN</em></u>...
Answer:
2S02 + O2 ---> 2SO3
<u><em>Its</em><em> </em><em>balanced</em><em> </em><em>now</em><em> </em><em>as</em><em> </em><em>Reactants</em><em> </em><em>=</em><em> </em><em>Products</em><em>.</em><em> </em></u>
write an equation to represent the oxidation of an alcohol.
identify the reagents that may be used to oxidize a given alcohol.
identify the specific reagent that is used to oxidize primary alcohols to aldehydes rather than to carboxylic acids.
identify the product formed from the oxidation of a given alcohol with a specified oxidizing agent.
identify the alcohol needed to prepare a given aldehyde, ketone or carboxylic acid by simple oxidation.
write a mechanism for the oxidation of an alcohol using a chromium(VI) reagent.
The reading mentions that pyridinium chlorochromate (PCC) is a milder version of chromic acid that is suitable for converting a primary alcohol into an aldehyde without oxidizing it all the way to a carboxylic acid. This reagent is being replaced in laboratories by Dess‑Martin periodinane (DMP), which has several practical advantages over PCC, such as producing higher yields and requiring less rigorous reaction conditions. DMP is named after Daniel Dess and James Martin, who developed it in 1983.
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
Cr2O2−7+14H++6e−→2Cr3++7H2O
Answer:
Number of moles of sodium reacted = 0.707 moles
Explanation:
P(H₂) = P(T) – P(H₂O)
P(H₂) = 754 – 17.5 = 736.5 mm Hg
Use the ideal gas equation which
PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature
<u>Re- arrange to calculate the number of moles and using the data provided</u>
n = P x V/R x T
n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)
n = 0.35348668
n = 0.353 moles H₂
<u>from the equation we know that</u>
0.353 mole H₂ x 2mole Na/1mole H₂, So
0.353 x 2 = 0.707 mole Na
The number of moles of Sodium metal reacted were 0.707 moles.
The balanced chemical reaction would be:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:
Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl
0.646 - 0.48 = 0.166 mol HCl left
