We have to calculate the percentage (by mass) of Ba in the mixture of BaBr₂ and inert material.
Given, bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate as per following reaction:
BaBr₂(aq) + 2AgNO₃(aq)→2AgBr(s)+Ba(NO₃)₂(aq). From this balanced chemical reaction, it is clear that one mole BaBr₂ reacts with two moles of AgNO₃ and gives two moles of AgBr. We have to calculate 0.6226 g AgBr contains how many moles of AgBr. Using that information we can get how many moles of BaBr₂ reacted to give 0.6226g AgBr and one mole BaBr₂ contains one mole of Ba. By multiplying number of moles of Ba with atomic mass of Ba we can get amount of Ba present in the mixture. Accordingly we can calculate mass percentage of Ba in the mixture.
Atomic mass of Ba= 137.327 g.mol, Molecular mass of BaBr₂=297.1 g/mol and molecular mass of AgBr=187.7 g/mol. Mass of AgBr is 0.6226 g which contains 0.6226/187.7 mole= 3.31X10⁻³ moles of AgBr. So, moles of BaBr₂ reacts= (3.31X10⁻³)/2 moles= 1.65X10⁻³ moles. One mole BaBr₂ contains one mole of Ba. So, 1.65X10⁻³ moles of BaBr₂ contains 1.65X10⁻³ moles of Ba atoms whose mass is=1.65X10⁻³X137.327g=0.2265 g. Out of 0.7207 g sample amount of Ba present is 0.2265 g. So, mass percentage is =31.42 %
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Answer:
A)The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
B)Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
C)If the product is wet with water there will be no change in the infrared spectrum
Explanation:
The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
If the product is wet with water there will be no change in the infrared spectrum
For the reactants,
- The oxidation number of hydrogen = +1
- The oxidation number of oxygen = -2
- The oxidation number of arsenic = +5
- The oxidation number of carbon = +3
For the products,
- The oxidation number of hydrogen = +1
- The oxidation number of oxygen = -2
- The oxidation number of arsenic = +3
- The oxidation number of carbon = +4
Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.
Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.
Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.
So, the answers are:
name of the element oxidized: Carbon
name of the element reduced: Arsenic
formula of the oxidizing agent:
formula of the reducing agent:
Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
If: ΔT (T final - T initial) = ?
In a redox chemical reaction, one species gets reduced and another gets oxidized. Manganese element is reduced in this reaction.
<h3>What is oxidized and reduced?</h3>
In a redox reaction, the increase or decrease in the oxidation number and electrons results in the reduction and oxidation of the chemical species. The oxidation and reduction occur simultaneously in a reaction.
The oxidation number of Mn in permanganate ion was +8 on the left side and decreased to +4 on the right side of the equation. Potassium permanganate is an oxidizing agent that has reduced the manganese ion of the permanganate ion.
Therefore, manganese is reduced.
Learn more about reduction and oxidation here:
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