The correct answer is C. Atoms are incredibly small and can bearly be seen with the most powerful electron microscopes. The nucleas of an atom contains protons and neutrons with electrons in orbitals around the nucleas. I hope this helps. Let me know if anything is unclear.
Answer:
B
Explanation:
B is the best showing of a chemical reaction out of the choices
I got that pH=3.65 using the fact that Ka=[H⁺][A⁻]/[HA] at equilibrium. In the ice table, I stands for initial, C stands for change, and E stands for equilibrium.
I hope this helps. Let me know if anything is unclear.
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!
Substituting the values:
51 + 3(131) = ΔH + 2(28) + 3(189)
ΔH = -225 J/mol
When written outside of the equation, this becomes 225 J/mol
