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 %
This is asking for the absolute humidity. So if the relative humidity is 65 % and the air temperature is 30 then we need to have in mind the barometric pressure which will be 760. The humidity in the air will be 0.020 kg/m3. In order to know this we calculate it eith the mass of water vapour in a unit volume of air. This is a measure of the actual water vapour content of the air. the for mula would go like this: AH = Mv / V.
Two chromate ions are contained in this compound. The “di” prefix means Two
Mono- one
Di- two
Tri- three
Tetra- four
Penta- five
Hexa-six
Hepta- seven
Octa- eight
Nona- nine
Deca- ten
NaHCO3 = No. of atoms are 1 sodium + 1 Hydrogen + 1 carbon + 3 oxygens = 6 atoms per molecule.
so a total of 3 oxygens
Answer:
Contact
Explanation:
As the speed of the particles increases, they move around more, which means they hit the sides of the container more.
