Draw two glycine molecules and show how they can be linked by a condensation reaction

Calculate the wavelength for the transition from n = 4 to n = 2, and state the name given to the spectroscopic series to which t

finlep [7]

Answer:

The wavelength for the transition from n = 4 to n = 2 is 486nm and the name name given to the spectroscopic series belongs to The Balmer series.

Explanation

lets calculate -

Rydberg equation- $\frac{1}{\pi } =R(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

where , $\pi$ is wavelength , R is Rydberg constant ( $1.097\times10^7$ ), $n_1$ and $n_2$ are the quantum numbers of the energy levels. (where $n_1=2 , n_2=4$ )

Now putting the given values in the equation,

$\frac{1}{\pi }=1.097\times10^7\times(\frac{1}{2^2} -\frac{1}{4^2} )$ $=2056875m^-^1$

Wavelength $\pi =\frac{1}{2056875}$

= $4.86\times10^-^7$ = 486nm

 Therefore , the wavelength is 486nm and it belongs to The Balmer series.

8 0

2 years ago

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

2 years ago

Draw the best Lewis structure for CH 3 -1. What is the formal charge on the C?

djyliett [7]

Answer : The formal charge on the C is, (-1) charge.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CH_3^{-1}$

As we know that carbon has '4' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in $CH_3^{-1}$ = 4 + 3(1) + 1 = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge on carbon atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

The formal charge on the C is, (-1) charge.

3 0

3 years ago

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 0

2 years ago

Anyone know the answer to number 1?

Andrew [12]

The rate at which the substance dissolves in water.

5 0

3 years ago