All categories

3 years ago

Determine what elements are denoted by the following electrons: [Kr] 5s2 4d10 5p3

Chemistry

1 answer:

RoseWind [281]3 years ago

4 0

Answer:

Antimony Sb

Explanation:

ntimony atoms have 51 electrons and the shell structure is 2.8.18.18.5.

The ground state electron configuration of ground state gaseous neutral antimony is [Kr].4d10.5s2.5p3

You might be interested in

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

Which chemical bond (be specific) in a nucleotide is broken to supply the energy for nucleic acid synthesis

Korvikt [17]

Answer:

Phosphodiester Bond

Explanation:

Nucleotides are linked with a phosphodiester bond formed between the phosphates group present in Nucleotides.

Nucleic acids including DNA and RNA in living organisms. Phosphodiester bond between nucleotide is broken to supply the energy for nucleic acid synthesis and supply energy between the 5' and 3' carbon atoms in Nucleic acids.

Hence, the correct answer is "Phosphodiester Bond".

7 0

3 years ago

For the reaction ? Ch3oh+? O2 →? Hco2h+? H2o , what is the maximum amount of hco2h (46.0254 g/mol) which could be formed from 6.

saw5 [17]

Answer:- 9.98 grams.

Solution:- The given balanced equation is:

$CH_3OH+O_2\rightarrow HCO_2H+H_2O$

From this equation, there is 1:1 mol ratio between methanol and formic acid. We start with given grams of methanol and convert them to moles. Then using mol ratio the moles of foramic acid are calculated and finally converted to grams on multiplying by it's molar mass.

The calculations could easily be done using dimensional analysis as:

$6.95gCH_3OH(\frac{1molCH_3OH}{32.0419g})(\frac{1molHCO_2H}{1molCH_3OH})(\frac{46.0254gHCO_2H}{1molHCO_2H})$