80.24% carbon, 9.62% hydrogen and 10.14% oxygen
1) Base: 100 g
C: 80.24 g
H: 9.62 g
O: 10.14 g
----------
100.00 g
2) Conversion to moles
C: 80.24 g / 12.00 g /mol = 6.687 mol
H: 9.62 g / 1.00 g/mol = 9.62 mol
O: 10.14 g / 16.00 g/mol = 0.63375 mol
3) Ratio
C:6.687 / 0.63375 = 10.55
H: 9.62 / 0.63375 = 15.18
O: 0.63375 / 0.63375 = 1
Given that the tenth of C is 0.5 you need to multiply all the numbers by 2 =>
C: 10.55 * 2 = 21.1--> 21
H: 15.18 * 2 = 30.36 -> 30
O: 1 * 2 = 2
=> C21 H30 O2
Answer: C21 H30 O2
The average atomic mass of tellurium, calculated from its eight isotopes (Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), Te-128 (31.79%), and Te-130 (34.48%)) is 127.723 amu.
The average atomic mass of Te can be calculated as follows:
Where:
m: is the mass
%: is the abundance percent
Knowing all the masses and abundance values, we have:
To find the <u>average atomic mass</u> we need to change all the <u>percent values</u> to <u>decimal ones</u>
Therefore, the average atomic mass of tellurium is 127.723 amu.
You can find more about average atomic mass here brainly.com/question/11096711?referrer=searchResults
I hope it helps you!
Moles= number of molecules/Avogadro’s constant
(3.8 x 10^24) /(6.02 x 10^23)
=6.31229..
Volume=moles*24(molar gas volume under STP)
= 6.31229...*24
= 151.5 dm^3 (units mainly used in the UK)
=. 151.5 litres
46-8= 38 minutes
It took him 38 minutes
Following the equation and the stoichiometry of the reaction, excess reactant.
<h3>What is excess reactant?</h3>
The excess reactant has to do with that reactant that is present in more amount than is required in the reaction.
In this case, we have the reaction; 2NBr3 + 3NaOH ---> N2 + 3NaBr + 3HOBr.
If 2 moles of NBr3 reacts with 3 moles of NaOH
x moles of NBr3 reacts with 48 moles of NaOH
x = 32 moles
Hence, NBr3 is the excess reactant.
Learn more about excess reactant: brainly.com/question/14449229
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