Answer:
atom are very tiny partical
A COVALENT BOND, FORMS BETWEEN ELEMENTS WITH SIMILAR ELECTRONEGATIVITY AS SHARING OF ELECTRON PAIRS BETWEEN ATOMS IS EASIER AS THEY ARE IDENTICAL.
Explanation:
Bonding atoms with similar electronegativity values form covalent bonds.
A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms.
Covalent bonds form between two nonmetal atoms with identical or relatively close electronegativity values
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons, also it is the strength an atom has to attract a bonding pair of electrons to itself.
Pure covalent bonds result when two atoms of the same electronegativity bond. This occurs only when two atoms of the same element bond with each other.
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16
Mr = 24 + 28 + 16 x 3
Mr = 100
moles = mass / Mr
moles = 237 / 100
moles = 2.37
Answer:
Explanation:
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