Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
Answer:
positron
Explanation:
A positron is a particle produced when a proton is transformed into a neutron. Anti neutrinos are ejected from the nucleus to balance spins.
Positron emission increases the Neutron/Proton ratio. When a nuclide undergoes positron emission, the atomic number of the daughter nucleus is one unit less than that of the parent nucleus hence it is found one place before its parent in the periodic table.
The balanced equation for the above reaction is as follows;
2Ca + O₂ ---> 2CaO
stoichiometry of Ca to O₂ is 2:1
we first need to find the limiting reactant
number of Ca moles - 6.84 mol
number of O₂ moles - 4.00 mol
if Ca is the limiting reactant
if 2 mol of Ca reacts with 1 mol of O₂
then 6.84 mol of Ca reacts with - 6.84 / 2 = 3.42 mol of O₂
this means that Ca is the limiting reactant and O₂ is in excess
therefore amount of CaO produced depends on amount of limiting reactant present
stoichiometry of Ca to CaO is 2:2
number of moles of Ca reacted = number of CaO moles formed
number of moles of CaO formed - 6.84 mol
answer is 6.84 mol