Step 1
The reaction is written and balanced:
4 Rb + O2 =>2 Rb2O
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Step 2
Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100
The actual yield is provided by the exercise = 39.7 g
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Step 3
Determine the limiting reactant. The molar masses are needed to solve this:
For Rb) 85.4 g/mol
For O2) 32 g/mol
Procedure:
4 Rb + O2 =>2 Rb2O
4 x 85.4 g Rb ----- 32 g O2
82.4 g Rb ----- X = 7.72 g O2 are needed
For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.
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Step 4
Determine the theoretical yield from the limiting reactant:
The molar mass Rb2O) 187 g/mol
Procedure:
4 x 85.4 g Rb ------ 2 x 187 g Rb2O
82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield
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Step 5
% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.
Answer: % yield = 44 %
Answer:
B
Explanation:
so you started out with 80
you have to cool it till 60
so it is decreasing which means you have to subtract
Note=if something is increasing or wanting to go up its addition but for this problem 60 is less then 80 so we are doing subtraction
80-60=20
Answer: D
Explanation:
A control group is the variable that remains constant throughout the whole experiment
Answer:
An emulsifying agent is typically characterized by having <u><em>d. one polar end and one nonpolar end.</em></u>
Explanation:
Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.
This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.
<u><em>An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.</em></u>
