<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:
B. Two chlorine atoms
Explanation:
This electronic configuration shows that the given atom is magnesium.
Electronic configuration of magnesium:
Mg¹² = 1s² 2s² 2p⁶ 3s²
There are two valance electrons of magnesium that's why it would react with two atoms of chlorine. Chlorine is present in seventeen group. It has seven valance electrons. It required just one electron to get complete octet. While magnesium needed to lose two electrons to get complete octet. That's why two chlorine atoms bonded with one magnesium atom. Thus both would get complete octet.
Mg + Cl₂ → MgCl₂
Answer:
NaOH is the limiting reactant.
Explanation:
Hello there!
In this case, since the reaction taking place between sodium hydroxide and chlorine has is:
Which must be balanced according to the law of conservation of mass:
Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:
Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.
Regards!
