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3 years ago

The specific rotation of (S)-2-butanol is +13.5. If 0.119 g of its enantiomer is dissolved in 10.0 mL of ethanol and

Chemistry

1 answer:

larisa86 [58]3 years ago

7 0

Answer:

+15.8°

Explanation:

The formula for the observed rotation (α) of an optically active sample is

α = [α]lc

where

l = the cell path length in decimetres

c = the concentration in units of g/100 mL

[α] = the specific rotation in degrees

1. Convert the concentration to units of g/100 mL

$c = \frac{\text{0.119 g}}{\text{10.0 mL}} \times \frac{10 }{10 } =\frac{\text{1.19 g}}{\text{100 mL}}$

2. Calculate the observed rotation

$\alpha = (+13.3 ^{\circ}) \times 1.19 \times 1 = +15.8^{\circ}$

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6.02 times 10 to the 23 power, which is Avogadros number

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3 years ago

The pKb value for aqueous ammonia at 25

Alex17521 [72]

The value of pka for $NH_3$ in an aqueous solution is 9.2.

Kb denotes the base dissociation constant.

pKa + pKb =14 at 25 degree celcius.

pKa + 4.8 =14

pKa = 9.2

pKb is the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.

It is used to determine the strength of a base or alkaline solution.

The value of pka for $NH_3$ in an aqueous solution is 9.2.

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1 year ago

Identify structure(s) for all constitutional isomers with the molecular formula C4H8 that have one double bond. Select all that

Alchen [17]

Answer:

a) But-1-ene

b) E-But-2-ene

c) Z-But-2-ene

d) 2-Methylpropene

Explanation:

In this case, if we want to draw the isomers, we have to check the formula $C_4_H8$ in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain But-1-ene.

For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain Z-But-2-ene. When the methyls groups are placed on opposite sides we will obtain E-But-2-ene.

Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain 2-Methylpropene.

See figure 1 to further explanations.

I hope it helps!

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2 years ago

The element magnesium has significantly fewer protons and neutrons than gold. If you were to use MgMg in place of AuAu in the Ru

Vadim26 [7]

Answer:The center will be quite smaller in magnesium than in gold thus the alpha particle will scatter to a lesser degree and the experiment may not yield a meaningful result.

Explanation:

Gold has seventy nine protons and magnesium has only twelve protons. Gold atoms are quite much more massive than magnesium atoms. This implies that if magnesium atoms were used in Rutherford's experiment, the number of protons will be too few to give a reasonable result since the atom is quite small. A larger atom, having a larger centre is required in order to have meaningful scattering of alpha particles.

5 0

3 years ago

Calculate the amount in moles and grams of KMnO4 present in 3dm^3 of 0.25M

Stels [109]

Answer:

Number of moles = 0.75 mol

Mass = 118.5 g

Explanation:

Given data:

Number of moles of KMnO₄ = ?

Mass in gram = ?

Molarity of KMnO₄ = 0.25 M

Volume of KMnO₄ solution = 3 dm³ (3 L)

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles / L of solution

0.25 M = number of moles / 3 L

Number of moles = 0.25 M× 3 L

M = mol/L

Number of moles = 0.75 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.75 mol × 158 g/mol

Mass = 118.5 g

5 0

2 years ago