Answer:
stars,auroras,a neon sign
Explanation:
just did it
Answer:
Group 7A
Explanation:
The group 7A elements consists of the most reactive non-metals on the periodic table.
This group is known as the group of halogens. They consist of element fluorine, chlorine, bromine, iodine and astatine.
- The elements in this group have the highest electronegativity values.
- They have 7 valence electrons and requires just one electron to complete their octets.
- This way, they are highly reactive in their search for that single electron.
<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D%7B%5B%5Ctext%7Bacid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
![[\text{conjuagate base}]}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D)
= moles of conjugate base = 3.25 moles
![[\text{acid}]](https://tex.z-dn.net/?f=%5B%5Ctext%7Bacid%7D%5D)
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 4.61
Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
