Answer:
K2SO4, NH3, HOCI, HCI, CH3NH2, SiCl4, CO2, CH20
Explanation:
Substances are soluble in water when they are ionic or polar covalent substances.
If we look at the substances listed, K2SO4 is ionic while NH3, HOCI, HCI, CH3NH2, SiCl4, CO2, CH20 all contain polar covalent bonds which accounts for their water solubility.
Hence ionic and polar covalent substances are soluble in a polar solvent such as water.
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)
Answer:
molarity = moles of solution/liters of solution
molarity = 1 mole/2 liters
molarity = 0.5 M
molarity = 50 moles/200 kg
molarity = 0.25 M
Explanation:
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3
2) Only single bonds => alkane => sufix ane
3) no substitutions
4) 5 carbons = > prefix penta.
Therefore, the name is pentane.
Answer:
52 da
Step-by-step explanation:
Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.
The i<em>ntegrated rate law for a first-order reaction </em>is
ln([A₀]/[A] ) = kt
Data:
[A]₀ = 750 mg
[A] = 68 mg
t_ ½ = 15 da
Step 1. Calculate the value of the rate constant.
t_½ = ln2/k Multiply each side by k
kt_½ = ln2 Divide each side by t_½
k = ln2/t_½
= ln2/15
= 0.0462 da⁻¹
Step 2. Calculate the time
ln(750/68) = 0.0462t
ln11.0 = 0.0462t
2.40 = 0.0462t Divide each side by 0.0462
t = 52 da
