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:
I think the answer is Toledo
The event which is most likely occurring in this scenario is effusion because there is a movement of a gas through a small opening into a larger volume and is denoted as option C.
<h3>What is Effusion?</h3>
This is referred to as the process in which a gas or a substance escapes from a container through a hole of diameter which is usually smaller.
The type of event which is most likely occurring is effusion because of the presence of the small holes in which the balls are made to pass through the center which is why option C was chosen.
Read more about Effusion here brainly.com/question/2097955
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The options are:
- diffusion because particles move from regions of high concentration to regions of low concentration.
- diffusion because particles move from regions of low concentration to regions of high concentration.
- effusion because there is a movement of a gas through a small opening into a larger volume.
- effusion because there is a movement of a gas through a large opening into a smaller volume
Answer:
204.73K
Explanation:
the formula : PV=nRT
n=4
P=5.6 atm
V=12 L
R=0.08206 L atm mol-1 K-1
T=?
So, if you plug it in, you will get:-
T=PV/nR
T=(5.6 atm)(12 L)/(4 mol)(0.08206 L atm mol-1 K-1)
T=204.73 K
hope this is correct!
Answer:
101 L
Explanation:
35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O
35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O
35.0 g KOH is the limiting reactant
