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rewona [7]
3 years ago
7

A solution of 400mg of optically active 2-butanol in 10 mL of water, placed in a 20 cm cell shows an optical rotation of 40o. Wh

at is the specific rotation of this compound?
Chemistry
1 answer:
labwork [276]3 years ago
3 0

Answer:

Specififc rotation [∝] = 0.5° mL/g.dm

Explanation:

Given that:

mass = 400 mg

volume = 10 mL

For a solution,

The Concentration = mass/volume

Concentration = 400/10

Concentration = 40 g/mL

The path length l = 20 cm = 2 dm

Observed rotation [∝] = + 40°

Specififc rotation [∝] = ∝/l × c

where;

l = path length

c = concentration

Specififc rotation [∝] = (40 / 2 × 40)

Specififc rotation [∝] = 0.5° mL/g.dm

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What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th
zaharov [31]

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

E = E^{\circ} - \dfrac{RT}{zF}\ln Q

(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}}

 

7 0
3 years ago
Pls help this a test for Google classroom
muminat

Occurs at a specific temperature

5 0
2 years ago
Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.
kakasveta [241]

Answer:

The volume will be 89.6875 ml

Explanation:

So to count this we will use a single proportion.

0.0640 mol - 1000 ml

5.74×10−3 mol - x ml

x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml

6 0
3 years ago
A sample of an unknown compound is vaporized at 150.°C . The gas produced has a volume of 960.mL at a pressure of 1.00atm , and
IrinaK [193]

Answer:

34.02 g.

Explanation:

Hello!

In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

PV=nRT\\\\n=\frac{1.00atm*0.960L}{0.08206\frac{atm*L}{mol*K}*(150+273)K} =0.0277mol\\\\

Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:

MM=\frac{0.941g}{0.0277mol} \\\\MM=34.02g/mol

So probably, the gas may be H₂S.

Best regards!

6 0
3 years ago
What are the names of the components (each shown with a leader and a line) of a circuit shown in the diagram? Please help!!!!
alekssr [168]
Maybe: Battery, wire, lamp and switch
6 0
3 years ago
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