It’s c I just did this so yea c
Answer:
Functional group
Explanation:
Functional group is specific group of atom or bond associated to an organic compound that determines the chemical properties of that compound. This atom is bonded in a certain way or specific arrangement to give the compound a peculiar physical and chemical characteristics.
Functional group like the alkyl group -CH3 is found in organic compound series like the alkane family. The chemical properties specific to the alkyl group will be active in the compound of alkane family because of the presence of the functional group Alkyl(-CH3). The functional group also plays a major role in the chemical reactivity of the compound. For example the functional group of alkyl are often non reactive, this non reactive nature will definitely rub off on the chemical properties of the compound it is attached.
Functional group like -OH is usually found in Alcohol . Due to the presence of this functional group (-OH) alcohol possess a peculiar chemical properties. The compounds possess an hydrogen bond which invariably lead to the higher boiling points of the alcohol compounds. Other functional group can be bonds like double bond found in alkene compound or triple bond found in alkyne compound.
If I made no mistake in calculation, the given answer must be correct...(tried my best)
elements : carbon hydrogen oxygen Fluorine
composition [C] 24 3 16 57
M r 12 1 16 19
(divide C by Mr) 2 3 1 3
(Divide by smallest value) 2 3 1 3
(smallest value = 1...so all value remained constant)
Empirical formula : C2H3OF3
if molar mas = 100 g per mole, then
first step calculate Mr. of empirical formula: [= 100]
Them molecular formula = empirical formula
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.
From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
However, this is the molarity of the HCl and NaOH solution, not the original HCl solution. Using the dilution equation M1V1=M2V2, we can solve for the original molarity.
M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
