Magnesium and calcium are 2nd group elements with 2 valence electrons. They are metals in room temperature and easily lose these electrons to nonmetals such as oxygen.
What is magnesium ?Magnesium is 12th element in periodic table. It is 2nd group element and is a called alkaline earth metals. Its group members are calcium, strontium, and barium.
They contains 2 valence electrons which can be easily lost to a non-metal. Oxygen is highly electronegative element and it contains 6 valence electrons and need two more electrons to achieve octet.
Al is 13th group element it is not as much electropositive as alkaline earth metals. It contains 3 valence electrons and thus it need to lose all these 3 electrons to achieve octet. Hence, Al is less reactive towards O2 in comparison with Mg and Ca.
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What makes a field forever unique?
From cotton fields to corn stalks, if this is what you mean, a forever field is unique in its own way by keeping all types of plants in the field. They are all different in how they grow and what they do.
A science class is going to do an experiment to investigate how water temperature changes the rate at which salt dissolves in water. What tool will be needed to measure the temperature of the water?
Answer:
thermometers
Explanation:
Answer: thermometer
Explanation: trust me ;)
Dna has an important role in making proteins, the biomolecules that help determine our traits as well as take part in chemical reactions throughout our bodies. Can you correctly label the steps/parts involved in protein synthesis?.
Protein synthesis is a process that is generated within cells where different organelles and genetic components are involved through which proteins are generated.
What are the steps of protein formation?The biosynthesis of proteins will be given by different parts, beginning with the translation of the genetic material. It will be given by different steps:
1. mRNA binds to ribosomes and the aminoacyl-tRNA to be associated to the first codon of the mRNA is associated.
2. The elongation of the polypeptide chain occurs, in which the amino acids are joined
3. The elongation is finished when the termination codons are reached.
4. The translation components are released and post-translational modifications are started.
5. The folding, glycosylation and modification of amino acids occur
6. Finally, the protein is released to generate its function
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ammonium ion (nh4 ) is an acid that can dissociate to form ammonia (nh3) and h in aqueous solution. the pka for this reaction is 9.25. when total ammonium ion is 50% dissociated, what is the ph?
When total ammonium ion is 50% dissociated, The pH is 9.25.
Given, the pKₐ of ammonium ion (NH₄⁺) is 9.25.
The dissociation of NH₄⁺ to ammonia (NH₃) and H⁺ in aqueous solution is represented below.
It is stated that the ammonium ion is 50% dissociated in the solution. Therefore the concentration of each species at that instant [if started with 1M conc. of NH4+ (aq)] is summarized as follows:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
Initial conc. (in M) 1 - -
At 50% dissociation 1-0.5, i.e., 0.5 0.5 0.5
The given value of pKₐ immediately reminds us of the form of Henderson-Hasselbalch equation containing pKₐ , i.e.,
pH = pKₐ + log₁₀ [tex]\frac{[salt]}{[acid]}[/tex] _ _ _(i)
But, NH₄⁺ (aq) exists as the conjugate acid form, as well as the salt (ionized) form. Hence, we cannot use equation. (i) for our calculations.
Therefore, we need to convert the above value of Kₐ to the equivalent Kb (for NH₃), corresponding to the reverse equilibrium.
This is achieved by using the relation
Kb = [tex]\frac{Kw}{Ka}[/tex]
or, -log₁₀ Kb = -log₁₀ Kw + log₁₀ Ka
or, pKb = pKw - pKa = 14 - 9.25 = 4.75
where, pKw = 1 x 10⁻¹⁴ (at 298 K).
Therefore, pKb of ammonia (NH₃) = 4.75
Now, we need to use the form of Henderson-Hasselbalch equation which uses Kb; which is in fact
pOH = pKb + log₁₀ [tex]\frac{[salt]}{[base]}[/tex] _ _ _(ii)
where the salt is NH₄⁺, and the base is NH₃. Further, we know that
Kw = [H⁺][OH⁻] or, pOH + pH = pKw = 14 or, pOH = 14 - pH
Substituting this relation in eq. (ii) gives
14 - pH = pKb + log₁₀ [tex]\frac{[NH4+]}{[NH3]}[/tex]
or, pH = 14 - pKb — log₁₀ [tex]\frac{[NH4+]}{[NH3]}[/tex]
As calculated, pKb = 4.75, [NH₄⁺] = 0.50 M, while [NH₃] = 0.50 M, assuming we start with a conc. of 1M NH₄⁺. Hence,
pH = 14 – 4.75 – log₁₀ [tex]\frac{0.50}{0.50}[/tex]
= 9.25 - log₁₀ ( 1 )
= 9.25 - 0
pH = 9.25
Hence, the pH of a solution consisting of 50% NH₃ and 50% NH₄⁺ will be equal to 9.25.
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