Answer:
Option C
CH₃CH₂CH₂COOH
Explanation:
Carbonxylic acids are compounds which has the general formula
R–COOH where R is an alkyl group.
Considering the options given in the question above,
For A:
CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.
For B:
CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.
For C:
CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.
For D:
CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group
For E:
ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.
From the above illustration, only option C contains a Carbonxylic compound.
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
Answer:
Explanation:
mole of HCl remaining after reaction with CaCO₃
= .3 M of NaOH of 32.47 mL
= .3 x .03247 moles
= .009741 moles
Initial HCl taken = .3 x .005 moles = .0015 moles
Moles of HCl reacted with CaCO₃
= .009741 - .0015 = .008241 moles
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O .
1 mole 2 moles
2 moles of HCl reacts with 1 mole of CaCO₃
.008241 moles of HCl reacts with .5 x .008241 moles of CaCO₃
CaCO₃ reacted with HCl = .5 x .008241 = .00412 moles
the mass (in grams) of calcium carbonate in the tablet
= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )
Answer:
the last answer is right.
Answer:
C) The compound is largely ionic with A as the cation.
Explanation:
Pulings proposed the method to determine if the compound is ionic in nature or covalent in nature , by finding the difference between the electronegativity of the respective cation and anion .
The ion with higher electronegativity is the anion and the ion with lower electronegativity is the cation.
The electronegativity difference above 1.7 make the compound ionic in nature.
Hence, from the question ,
A is the cation and B is the anion.
And the electronegativity difference above 1.7 so the compound is ionic in nature.
