Answer:
Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Explanation:
1 mole of any compound contains 6.023 × 10²³ molecules.
molecular weight of NaCl is 23 + 35.5 = 58.5 g.
so, 58.5 grams of NaCl makes 1 mole
⇒ 14.5 g of NaCl = 
= 0.248 moles.
⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules
= 1.49 × 10²³ molecules.
And 1 molecule contains 1 Na ion and 1 Cl ion.
⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Answer:
Double replacement reaction
Explanation:
Now, let us first write the reaction equation properly:
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
The above reaction is a neutralization reaction between an acid and a base whose product gives salt and water only at most instances.
From here, we can observe that the species displaces on another in their ionic state. Hydrogen replaces potassium and water is produced. Potassium combines chemically with sulfate ions to give the salt of potassium.
It obtains a neutral pH (7). The combination of a strong acid and a strong base results in a neutral pH. The pH of HCl is about 1, and the pH of NaOH is 14.
Answer:
CuCl2-Ion-dipole forces
CuSO4-Ion-dipole forces
NH3-Dipole-dipole forces
CH3OH-Dipole-dipole forces
Explanation:
Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.
Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.
Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles
If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation. (Option C)
<h3>How do systems at equilibrium respond to perturbation?</h3>
When a system at equilibrium suffers a perturbation, it shifts its equilibrium position to counteract such perturbation.
Let's consider a solution of acetic acid at equilibrium.
CH₃CO₂H(aq) = CH₃CO₂⁻(aq) + H⁺(aq)
If more acetic acid were added to the solution, the system will shift toward the products to counteract such an increase.
How would the system change if more acetic acid were added to the solution?
A. [H⁺] would decrease and [CH₃CO₂⁻] would increase. NO.
B. [H⁺] and [CH₃CO₂⁻] would decrease. NO.
C. [H⁺] and [CH₃CO₂⁻] would increase. YES. Both products would increase.
D. [H⁺] would increase and [CH₃CO₂⁻] would decrease. NO.
If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation.
Learn more about equilibrium here: brainly.com/question/2943338
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