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mash [69]
1 year ago
9

What is the electron configuration for magnesium (Mg)?

Chemistry
1 answer:
AveGali [126]1 year ago
6 0

Magnesium has atomic no 12

Electronic configuration given by

  • 1s²2s²2p⁶3s²

Or

  • [Ne]3s²
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2 years ago
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DNA transcription-to-translation # 1 Homework Unanswered Due in 4 days Given the following sequence of the coding strand, writte
uysha [10]

Explanation:

Translation is the process by which a polypeptide is polymerized from genetic information.

Firstly we have to make a transcription from the coding DNA strand to a single RNA strand (mRNA). RNA pol reads from 5' to 3' of the template strand and nucleotides are added by complementarity ( Adenine with Uracil, Thymine with Adenine and Cytosine with Guanine, Guanine with Cytosine).

DNA:  5'-  CGTTATGTGGACTCTCTGGTATGACTCACCTTAT -3'

mRNA: 5'-GCAAUACACCUGAGAGACCAUACUGAGUGGAAUA -3'

mRNA goes to the ribosomes where translation takes place. The enzyme will read every three letters (codon) starting at the start codon sequence (TAC in DNA, AUG in mRNA). According to codons tRNA carrying the amino acids will place it (by complementary to their anticodon) and the enzyme will join it to the nascent polypeptide or protein.

In order to do this we need to look up the genetic code and assign the proper amino acids.

Unfortunately the given strand does not have a start codon TAC codifying for initial methionine.

3 0
3 years ago
The blank solution used to calibrate the spectrophotometer is 10.0 mL of 0.2 M Fe(NO3)3 diluted to 25.0 mL with 0.1 M HNO3. Why
Sliva [168]
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4 0
2 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
2 years ago
You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?
Rudiy27

Answer:

amount of silver chloride required is 0.015 moles or 2.1504 g

Explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

                                                     =143.36 g/mol

mass of AgCl = moles *molar mass

                       =0.015*143.36

                        =2.1504g

                        =

4 0
2 years ago
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