All categories

2 years ago

Relate the spectral lines to energy levels in an atom and the term quanta

1 answer:

4 0

Relate spectral lines to energy levels in an atom and to the term quanta. When atoms absorb energy and are in their excited state, their electrons “jump” to higher energy levels. The electrons have to lose this energy in order to “jump” back to their original energy level, and they do this by emitting light.

You might be interested in

Please help!!

Phantasy [73]

1.
cost in cents

cost = 5m

2.
miles = c + 2

7 0

2 years ago

Read 2 more answers

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

2 years ago

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po

SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide ( $C_{7}H_{7}NO$ ) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is $2.7^{o}C$ lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride ( $NH_{4}Cl$ ) are dissolved in the same mass of X, the freezing point of the solution is $9.9^{o}C$ lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

Moles of benzamide = $\frac{mass}{\text{Molar mass of benzamide}}$

= $\frac{70.4 g}{121.14 g/mol}$

= 0.58 mol

Now, we will calculate the molality as follows.

Molality = $\frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}$

= $\frac{0.58 mol}{0.85 kg}$

= 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

dT = $i \times K_{f} \times m$ ,

where, dT = change in freezing point = $2.7^{o}C$

i = van't Hoff factor = 1 for non dissociable solutes

$K_{f}$ = freezing point constant of solvent

m = 0.6837

Therefore, putting the given values into the above formula as follows.

dT = $i \times K_{f} \times m$ ,

$2.7^{o}C = 1 \times K_{f} \times 0.6837 m$

$K_{f}$ = 3.949 C/m

Now, we use this $K_{f}$ value for calculating i for $NH_{4}Cl$

So, moles of ammonium chloride are calculated as follows.

Moles of $NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}$

= 1.316 mol

Hence, calculate the molality as follows.

Molality = $\frac{1.316 mol}{0.85 kg}$

= 1.5484

It is given that value of change in temperature (dT) = $9.9^{o}C$ . Thus, calculate the value of Van't Hoff factor as follows.

dT = $i \times K_{f} \times m$

$9.9^{o}C = i \times 3.949 C/m \times 1.5484 m$

i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

2 years ago

In a study of the decomposition of the compound XX via the reaction

Leya [2.2K]

Answer:

( About ) 0.03232 M

Explanation:

Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [ $X_o$ ]"

This formula would be true for the following information -

{ $X_o$ = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }

________

Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6

1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X

1 / [X] = 28.8 + 1 / ( 0.467 ),

1 / [X] = 28.8 + 2.1413...,

1 / [X] = 31,

[X] = 1 / 31 = ( About ) 0.03232 M

Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M

3 0

2 years ago

Which air mass has formed immediately north of Antarctica in the image?

hammer [34]

Answer:

Maritime Polar

5 0

2 years ago

Read 2 more answers