Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
The answer to your question is: letter E
Explanation:
A. This option is correct, the n = 3 shell only has subshells: s, p and d, and shell n = 4 or 5 have f subshell.
B. This option is true in subshell p could be at most 6 electrons and 3 suborbitals.
C. This option is correct orbital "s" is a sphere.
D. This option is correct, in subshell d could be at most 10 electrons and 5 orbitals.
E. This option is false, hydrogen only has 1 electron and then one subshell (s).
Answer:
20 km north that's the answer hope it helped
<span>H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-
When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant.
when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....</span>
Answer: Sulphur, Phosphorus, Wood are non-conductors of electricity due to no presence of free electrons in outermost shell.
Explanation:
