To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16
Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.
C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96
Then add them up.
72.06+ 12+ 96= 180.06
Now find the percent composition of carbon.
72.06/ 180.06 x 100= 40.01%
So the answer is C 40%.
Answer:
THE CORRECT ANSWER IS COMMAND.HOPE IT HELP YOU
There's 6.022×10^23 particles in 1 mole of anything
like there is 1000 grams in 1 kilogram of anything
a) To find the mass after t years:we will use this formula:
A = Ao / 2^n when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)b) Mass after 90 y :by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mgC) Time for 1 mg remaining:when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
The equivalency point is at the point of the titration where the amount of titrant added neutralize the solution. When it’s a strong acid strong base titration, the equivalence point will be 7. When it is a weak acid strong base, the equivalence point it more basic (the exact number depends on what acid and base you use). And when it is a strong acid weak base, the equivalence number is more acid (the exact number depends on what acid and base you use). Hope this helps!