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3 years ago

In the molecule of SO2 there are two pie bonds these are formed Due to

1 answer:

4 0

These are formed when, after formation of the first atom, the pi-orbitals of Sulfur and Oxygen align and share another pair of electrons. This occurs between sulfur and both of the oxygen atoms, with each oxygen atom forming one pi-bond.

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What term describes a carbon atom that has four different groups bonded to it?

oee [108]

A carbon which is attached to four different atoms or group of atoms with different environment is called as Chiral Carbon or Asymmetric Carbon.

Non-<span>superimposable:
</span> The mirror image (molecule) of chiral carbon cotaining compounds are Non.Superimposable on each other. They are called enantiomers of each other.

Polarized Light and Chiral Carbon:
When a polarized light is allowed to fall on either enantiomer of chiral compound, it is rotated other clockwise or anti-clockwise.

Examples:
Below are three axamples of compounds containing chiral carbon.

8 0

2 years ago

28. Which set of coefficients will balance this chemical equation?

Makovka662 [10]

2021 hoped I helped you god bless you

5 0

3 years ago

You have a 28.2-g sample of a metal heated to 95.2°c. you drop it in a calorimeter with 100. g of water at 25.1°c. the final tem

Vlad [161]

The heat lost by the metal should be equal to the heat gained by the water. We know that the heat capacity of water is simply 4.186 J / g °C. Therefore:

100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp * (95.2°C - 31°C)

4 0

3 years ago

Convert 0.00200 g into Mg and answer in scientific notation:

Masja [62]

Answer: $2.00x10^{2}$

Explanation: 1 gram is equal to 1000 Milligrams so

$0.00200g=\frac{1000mg}{1g}$

Since you go straight across its just multiplication so 0.00200x1000=200mg

To answer it in scientific notation it should just be $2.00x10^{2}$ mg

5 0

1 year ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago