Answer:ionic
Explanation: just did it in edgenuity:)
Answer:
0.375 L
Explanation:
We know that at neutralization, the number of mol of acid must equal the number of equivalents of base.
This is a reaction 1:1 acid to base:
HClO₄ + NaOH ⇒ NaClO₄ + H₂O
We re given the moles of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.
Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol
Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:
Molarity = # moles / V ⇒ V = # moles / M
V = 0.028 mol / 0.0748 mol/L = 0.375 L
Note that this problem can be solved in just one step since
M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH) ⇒
V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)
Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
<u> E°/V</u>
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
<u> E°/V</u>
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41
</u>
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
</u>
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
![Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BFe%7D%5E%7B2%2B%7D%5D%5E%7B3%7D%7D%7B%20%5Ctext%7B%5BCr%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B0.25%5E%7B3%7D%7D%7B%200.75%5E%7B2%7D%7D%20%3D%5Cdfrac%7B0.0156%7D%7B0.562%7D%20%3D%200.0278%5C%5C%5C%5CE%20%3D%200.33%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B6%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%280.0278%29%5C%5C%5C%5C%3D0.33%20-0.00428%20%5Ctimes%20%28-3.58%29%20%3D%200.33%20%2B%200.0153%20%3D%20%5Ctextbf%7B0.35%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%5Clarge%5Cboxed%7B%5Ctextbf%7B0.35%20V%7D%7D)
You need to use the formula--> P1V1= P2V2 (Boyles's law)
P1= 14 bar
V1= 312 mL
P2= ?
V2= 652 mL
now we plug the values into the formula.
(14 x 312) = (P2x 652)
P2= (14 x 312)/ 652= <span>6.70 bar</span>
We can skip option B and D because NaCl is salt and H₂SO₄ is a strong acid.
Neutralization reactions are those reactions in which acid and base react to form salt and water.
As water being amphoteric in nature can react with HCl as follow,
HCl + H₂O ⇆ H₃O⁺ + OH⁻
In this case no salt is formed, so we can skip this option.
Ammonia being a weak base can abstract proton from HCl as follow,
HCl + NH₃ → NH₄Cl
Ammonium Chloride is a salt. So, among all four options, Option-C is the correct answer.
