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3 years ago

How many chiral centers does clavulanic acid have?

Chemistry

1 answer:

VikaD [51]3 years ago

7 0

Answer:

Clavulanic acid has two (2) chiral centers.

Explanation:

A chiral center is a center (usually carbon) with four different substituents.

The structure of clavulanic acid is shown in the attachment below.

Consider the labeled diagram in the attachment,

Carbon A is not a chiral carbon because it has two hydrogen atoms attached to it

Carbon B is not a chiral carbon because it has only three substituents

Carbon C is a chiral carbon because it has four different substituents

Carbon D is a chiral carbon because it has four different substituents

Carbon E is not a chiral carbon because it has only three atoms directly attached to it

Carbon F is not a chiral carbon because it has only three atoms directly attached to it

Carbon G is not a chiral carbon because it has two hydrogen atoms attached to it

Carbon H is not a chiral carbon because it has only three substituents

Then, only carbons C and D are chiral carbons.

Hence, clavulanic acid have two (2) chiral centers.

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Iron fillings sprinkled near a magnet arrange themselves into a pattern that illustrates the

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the force of attraction of iron fillings towards magnet

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3 years ago

g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30

Mekhanik [1.2K]

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol

3 0

4 years ago

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago

Find the ratio v/cv/c for an electron in the first excited state (n = 2) of hydrogen.

dezoksy [38]

The answer is 0.365:100.

v/c ratio represents ratio of speed of an electron (v) to the speed of light (c).

How is the speed of an electron calculated?

The speed of an electron (v) is given by Bohr's model as-

$v =\frac{1}{n}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}$

Now, for the first excited state, n = 2.

e - Charge of electron = $1.6$ × $10^{-19}$ C

h - Plank's constant = $6.6$ × $10^{-34} J.s$

ε₀- permittivity $= 8.85$ × $10^{-12}$ $N^{-1}.C^{2}.m^{-2}$

Put the above data in the formula-

$v =\frac{1}{2}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h} =\frac{1}{4}\times \frac{(1.6\times 10^{-19})^{2}}{8.85\times 10^{-12}\times 6.6\times 10^{-34}}\\ \\ =0.01096\times 10^{8} \\ \\=1.096\times 10^{6}ms^{-1}$