The empirical formula is the same as the molecular formula : C₁₀H₅O₂
<h3>Further explanation</h3>
Given
Molecular formula : C₁₀H₅O₂
Required
The empirical formula
Solution
The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.
The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
<em>(EF)n=MF
</em>
(EF)n = C₁₀H₅O₂
If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula
Answer:
1 kg
Explanation:
prefix nano means 10^-9
kilo 10^3
mili 10^-3
centi 10^-2
the largest amount is 1 kg
Answer:
Non competitive inhibition
Explanation:
Hello,
During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.
Best regards.
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.