The answer for this question would be B) False or the second option.
Answer is FALSE: ✅
Yes this is true because they are able to consume 2 organisms which are plants and animal just like us.
Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
