A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl dissolves. Some solid KCl is still visible in the flask. The solution in the flask is A) saturated and is at equilibrium with the solid KCl B) saturated and is not at equilibrium with the solid KCl C) unsaturated and is at equilibrium with the solid KCl D) unsaturated and is not at equilibrium with the solid KCl

Answer: The correct option is E.

Explanation: The reaction between aspirin (also known as acetylsalicylic acid) and sodium hydroxide is known as acid-base titration reaction.

By applying Unitary method, we get:

15.50mL of NaOH dissolves = 0.325 g of aspirin

So, 16.25 mL of NaOH will dissolve = = 0.341 g

Hence, the correct option is E.

Any member of the family of chemicals known as coordination compounds has a core metal atom that is surrounded by nonmetal atoms or groups of atoms, known as ligands, that are connected to it by chemical bonds. The name of the compound is tetraaminodiaquanickel (II)nitrate.

The additional molecular compounds known as coordination compounds are those that are stable in both the solid and dissolved states. In these compounds, ions or molecules connected by coordinate bonds connect the main metal atom or ion.

Coordination compounds are used in both vital catalytic processes that lead to the polymerization of organic molecules like polyethylene and polypropylene as well as hydrometallurgical processes that remove metals like nickel, cobalt, and copper from their ores.

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Answer:

pH at the equivalence point is 8.6

Explanation:

A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:

mmoles acid = mmoles of base

60 mL . 0.1935M = 0.2088 M . volume

(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH

The neutralization is:

HBz + KOH  ⇄  KBz  +  H₂O

In the equilibrum:

HBz + OH⁻   ⇄  Bz⁻  +  H₂O

mmoles of acid are: 11.61 and mmoles of base are: 11.61

So in the equilibrium we have, 11.61 mmoles of benzoate.

[Bz⁻] = 11.61 mmoles / (volume acid + volume base)

[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M

The conjugate strong base reacts:

  Bz⁻  +  H₂O  ⇄  HBz + OH⁻    Kb

0.1 - x                       x        x

(We don't have pKb, but we can calculate it from pKa)

14 - 4.2 = 9.80 → pKb  → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb

Kb = [HBz] . [OH⁻] / [Bz⁻]

Kb = x² / (0.1 - x)

As Kb is so small, we can avoid the quadratic equation

Kb =  x² / 0.1 → Kb . 0.1 = x²

√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M

From this value, we calculate pOH and afterwards, pH (14 - pOH)

- log [OH⁻] =  pOH → - log 3.98 ×10⁻⁶  = 5.4

pH = 8.6

Final answer:

To calculate the pH at equivalence in a titration, we need to consider the concentration of the excess strong base in the solution. First, we calculate the moles of the acid and the base, then we find the moles of the excess base. Using this information, we can find the concentration of the excess base and subsequently calculate pOH. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Explanation:

pH at the equivalence point in a titration can be determined by considering the concentration of the excess strong base present in the reaction mixture. In this case, the excess strong base is KOH. We can calculate [OH-] using the stoichiometry of the reaction and the given concentrations. Then, we can find the pOH using the formula -log[OH-]. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Given:

• Volume of benzoic acid solution (HC (H5CO2)): 60.0 mL
• Concentration of benzoic acid solution: 0.1935 M
• Concentration of KOH solution: 0.2088 M

Step 1: Determine the amount of benzoic acid (HC (H5CO2)) in moles:

moles of HC (H5CO2) = volume (L) × concentration (M) = 0.0600 L × 0.1935 M = 0.01161 mol

Step 2: Determine the amount of KOH in moles:

moles of KOH = volume (L) × concentration (M) = 0.0600 L × 0.2088 M = 0.01253 mol

Step 3: Determine the amount of excess KOH in moles:

moles of excess KOH = moles of KOH - moles required for neutralizing HC (H5CO2) = 0.01253 mol - 0.01161 mol = 9.2 × 10-4 mol

Step 4: Determine the concentration of excess KOH:

concentration of excess KOH = moles of excess KOH / volume (L) = 9.2 × 10-4 mol / 0.0600 L = 0.0153 M

Step 5: Determine the pOH of the solution:

pOH = -log[OH-] = -log(0.0153) ≈ 1.82

Step 6: Determine the pH of the solution:

pH = 14 - pOH = 14 - 1.82 ≈ 12.18

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Answer:

[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ;  [SO4⁻²] = 4,75x10⁻⁷M

SO4⁻²: 0.045ppm  ;  K⁺: 0.037ppm

[SO4⁻²] = 4,70x10⁻⁷ F

Explanation:

Determine the equation

K2SO4 → 2K⁺  +  SO4⁻²

Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion

Molar mass K2SO4: 174.26 g/m

Moles of K2SO4: grams / molar mass

2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles

Molarity: Moles of solute in 1 L of solution

1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)

K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M

SO4⁻²: 4,75x10⁻⁷ M

1 mol of K2SO4 has 2 moles of K and 1 mol of SO4

1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.

1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)

2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)

These grams are in 2.5 L of water, so we need μg/mL to get ppm

2.5 L = 2500 mL

1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)

9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)

113.35 μg /2500 mL = 0.045ppm

92.6 μg /2500 mL = 0.037ppm

Formal concentration of SO4⁻² :

Formality = Number of formula weight of solute / Volume of solution (L)

(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F

Answer:

Carbon (i) : quaternary carbon

Carbon (ii) : secondary carbon

Carbon (iii) : tertiary carbon

Carbon (iv) : secondary carbon

Explanation:

Carbons can be classified into 4 categories:

(1) Primary carbon: These are the atoms where the carbon atom is attached to one other carbon atom.

(2) Secondary carbon: These are the atoms where the carbon atom is attached to two other carbon atoms.

(3) Tertiary carbon: These are the atoms where the carbon atom is attached to three other carbon atoms.

(4) Quaternary carbon: These are the atoms where the carbon atom is attached to four other carbon atoms.

In the given structure:

Carbon (i) is attached to 4 further carbon atoms and hence, it is a quaternary carbon.

Carbon (ii) is attached to 2 further carbon atoms and hence, it is a secondary carbon.

Carbon (iii) is attached to 3 further carbon atoms and hence, it is a tertiary carbon.

Carbon (iv) is attached to 2 further carbon atoms and hence, it is a secondary carbon.