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3 years ago

8

About how many elements are found on the periodic table? Question 8 options:

1 answer:

krek1111 [17]3 years ago

3 0

the answer is actually 118 look it up

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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<u> 486nm</u> and the name name given to the spectroscopic series belongs to <u>The Balmer series.</u>

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

<u> Therefore , the wavelength is 486nm and it belongs to The Balmer series.</u>

8 0

3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago

in a sample of silcon 92.21% of the atoms have a mass of 27.98 amu, 4.70% have a mass of 28.98 amu and 3.09% have a mass of 29.9

oksano4ka [1.4K]

Answer: 28.1 amu

Explanation:

Mass of isotope 1 = 27.98 amu

% abundance of isotope 1 = 92.21% = $\frac{92.21}{100}=0.9221$

Mass of isotope 2 = 28.98 amu

% abundance of isotope 2 = 4.70% = $\frac{4.70}{100}=0.047$

Mass of isotope 3 = 29.97 amu

% abundance of isotope 2 = 3.09% = $\frac{3.09}{100}=0.0309$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(27.98 )\times 0.922+(28.98)\times 0.047+(29.97)\times 0.0309]$

$A=28.1amu$

Therefore, the average atomic mass of silicon is 28.1 amu

5 0

3 years ago

Which process transfers heat from a hot pavement to a person sitting on it?

Gennadij [26K]

conduction - the transfer of heat to an object through contact

6 0

4 years ago

Read 2 more answers

A student wants to make 5.00% solution of rubidium chloride using 0.37 g of the substance. what mass of water will be needed to

chubhunter [2.5K]

Answer:-

7.03 g

Explanation:-

Let the mass of water needed be M.

So total mass = M + 0.37

Now 5% of this solution has 0.37g of rubidium chlorude.

∴ (M +0.37) x (5/100) = 0.37

M +0.37 = 0.37 x 100/5

M + 0.37 = 7.4

M = 7.4-0.37

=7.03

Thus water to be added is 7.03 gram

7 0

3 years ago