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.
A. true
b. false
c. true
d. false
e. false
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
Answer: Sexual
Explanation: Two parents produce an offspring with some traits from each parent
