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Anni [7]
3 years ago
7

Let’s say that you have a solvatochromic compound that appears red in a solvent. You dissolve the compound in another solvent an

d it changes to an orange color. Would this be considered a hypsochromic or bathochromic shift? Would the solvent be more polar or less than the first solvent? Explain
Chemistry
1 answer:
EastWind [94]3 years ago
7 0

Answer:

Hypsochromic compound, More polar solvent

Explanation:

Hypsochromic shift refers to the shift of solution colour to blue side of the visible spectrum (blueshift) with increasing polarity of the solvent. In our case, the solution changes to orange colour from red when solvent is changed. This means that the emission spectrum of the solution underwent blueshift. (As orange colour is on the 'blue' side for red colour.) So this is a hypsochromic shift, and the new solvent is more polar that the previous one, as it caused hypsochromic shift.

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Relative to electrons and electron states, what does each of the four quantum numbers specify
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(n, l, m sub l, m sub s)

N: principle quantum number (1,2,3,4,etc)

l: angular momentum quantum number, the shape (l has to be at least 1 less than n, but can be 0 depending on n)

M sub l: magnetic quantum number (l determines this number)

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7. If all of the forces on an object are balanced, the object?
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3 years ago
A mole of oxygen and a mole of hydrogen (at STP) have all of the following in common EXCEPT
Drupady [299]

Answer:

Root mean squared velocity is different.

Explanation:

Hello!

In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

v_{rms}=\sqrt{\frac{3RT}{MM} }

Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.

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8 0
3 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act
algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y =  -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³  and  not 2.228 x 10⁻³). You can  recognize this mistake if you plot the data and notice it produces an outlier.

5 0
3 years ago
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