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STALIN [3.7K]
3 years ago
7

From where in the solar system did scientists conduct their spectral analyses in

Chemistry
1 answer:
katovenus [111]3 years ago
7 0

Answer:

Both Earth and Space

Explanation:

Because in 1948 there was no ability to analyze from space but in 1966 there was.

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Molecular polarity has a direct effect on the behavior of molecular compounds. The attractive forces between polar molecules are
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Stronger

Greater

Higher

Explanation:

Molecules are held together by intermolecular forces. These are forces that act between molecules in a particular state matter. Intermolecular forces depend on the nature of the molecule.

For polar molecules, the intermolecular forces are stronger thus it takes more energy to separate them leading to a higher boiling point of polar molecules irrespective of their molecular mass.

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How many moles are in 564 grams of Copper
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What is the chemical equation for a reaction between copper (ll carbonate and heat?
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3 years ago
Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i
anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

K=\frac{Products}{reagents}

In any case  

aA +Bb  equilibrium Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})

Where: H is reaction enthalpy, R is the gas constant and T temperature.  

Clearing the equation for K_{2} we get:

K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

Here we can study two cases: when delta H^{o} is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so K_{2} is greater that K_{1} and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so K_{2} will be less than K_{1} the forward reactions is not favored.  

{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

5 0
3 years ago
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