Answer: The IUPAC name of propan-2-one and is pentan-2-one.
Explanation:
The basic rules for naming of organic compounds are :
1. First select the longest possible carbon chain. The longest possible carbon chain should include the carbons of double or triple bonds or functional group.For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
2. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and ketones by adding the suffix -one.;
3. The numbering is done in such a way that the carbon containing the functional group gets the lowest number. While writing the name, the word root is given first , followed by the position of carbon containing the functional group followed by the suffix.
The IUPAC name of propan-2-one and is pentan-2-one.
The given question is incomplete. The complete question is:
A reversible reaction is said to have reached equilibrium when which of the following conditions is established?
a. concentration of reactants and products are equal
b. opposing reactions cease
c. speeds of opposite reactions become equal
d. temperature of opposite reactions become equal
Answer: c. speeds of opposite reactions become equal
Explanation:
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.The state of equilibrium refers to the dynamic state as both forward and backwad reactions continue.
For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equals to rate of the backward reaction.
Thus at chemical equilibrium, the amount of product and reactant remains constant because the rates of the forward and reverse reactions are equal.
Sodium hydrogen carbonate is a weak base because it is not a water soluble hydroxide.
Answer:
C. It does not participate in a decay series.
Explanation:
From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.
- It could have emitted any form of radioactive particles which can be alpha or beta.
- We do not know if it has a long or short half life because the value is not given.
- But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
- A decay series involves a radioactive decay in multiple steps.