Write a double replacement precipitation reaction A precipitate forms when aqueous solutions of nickel(II) chloride and silver(I
1 answer:
<u>Answer:</u> The balanced chemical equation is written below.
<u>Explanation:</u>
Double displacement reaction is defined as the reaction in which exchange of ions takes place.
Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.
When nickel (II) chloride reacts with silver (I) nitrate, it leads to the formation of white precipitate of silver chloride and an aqueous solution of nickel (II) nitrate.
The balanced chemical equation for the above reaction follows:
By Stoichiometry of the reaction:
1 mole of aqueous solution of nickel (II) chloride reacts with 2 moles of aqueous solution of silver (I) nitrate to produce 2 moles of solid silver chloride and 1 moles of aqueous solution of nickel (II) nitrate
Hence, the balanced chemical equation is written above.
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In the periodic table, elements of the same group are characterized by having the same similar properties.
So, first we will check the elements that lie within the same group as <span>beryllium and then we will attempt to choose the elements with atomic mass higher than 130.
So, elements in the same group as </span>beryllium are: magnesium, calcium, strontium, barium and radium.
Among these elements, we will find that:
radium has atomic mass of 226 amu
barium has atomic mass of 137.327 amu
Based on this, the two elements would be barium and radium.
So, first we will check the elements that lie within the same group as <span>beryllium and then we will attempt to choose the elements with atomic mass higher than 130.
So, elements in the same group as </span>beryllium are: magnesium, calcium, strontium, barium and radium.
Among these elements, we will find that:
radium has atomic mass of 226 amu
barium has atomic mass of 137.327 amu
Based on this, the two elements would be barium and radium.
Answer:
NaH₂PO₄ = 1.876 g
Na₂HPO₄ = 4.879 g
Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄
Explanation:
To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198
The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:
pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:
7.540 = 7.198 + log[A⁻]/[HA]
log[A⁻]/[HA] = 0.342
[A⁻]/[HA] =
[A⁻]/[HA] = 2.198
[A⁻] = 2.198*[HA]
The concentration of the acid and it's conjugated base must be equal to the concentration of the buffer 0.0500 M, so:
[A⁻] + [HA] = 0.0500
2.198*[HA] + [HA] = 0.0500
3.198*[HA] = 0.0500
[HA] = 0.01563 M
[A⁻] = 0.0500 - 0.01563
[A⁻] = 0.03436 M
The mix reaction is
NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻
The second ionization is:
H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺
So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).
So, the number of moles of these salts are:
NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol
Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol
The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:
NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol
Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol
The mass is the number of moles multiplied by the molar mass, so:
NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g
Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g
To prepare this buffer, it's necessary to have in solution the species H₂PO₄⁻ and HPO₄⁻², so it can be prepared for mixing the combination of:
H₃PO₄ and Na₂HPO₄ (the acid is triprotic so, it will form the H₂PO₄⁻ , and the salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);
H₃PO₄ and Na₃HPO₄ (same reason).
The other combinations will not form the species required.