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Roman55 [17]
2 years ago
8

Strong acids are assumed 100% dissociated in water. True As a solution becomes more basic, the pOH of the solution increases. Fa

lse The conjugate base of a weak acid is a strong base. True The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion. True As the Kb value for a base increases, base strength increases. True The weaker the acid, the stronger the conjugate base.
Chemistry
1 answer:
Kruka [31]2 years ago
7 0

Answer:

Strong acids are assumed 100% dissociated in water- True

As a solution becomes more basic, the pOH of the solution increases- false

The conjugate base of a weak acid is a strong base- true

The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion- True

As the Kb value for a base increases, base strength increases- true

The weaker the acid, the stronger the conjugate base- true

Explanation:

An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.

The pOH decreases as a solution becomes more basic (as OH^- concentration increases).

Ka refers to the dissociation of an acid HA into H3O^+ and A^-.

The greater the base dissociation constant, the greater the base strength.

The weaker an acid is, the stronger , its conjugate base will be.

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8 0
2 years ago
Select the correct layer.<br> In which layer is lithification most likely to occur?
Vera_Pavlovna [14]

Answer:

"Subsoil" is the correct answer.

Explanation:

  • The method of transforming sediments become rocky solids is considered as Lithification. The floor stratum or level below the upper floors at the base of the floor is a Subsoil.
  • Sediments comprise materials particles like sand, pebbles, skeletons kind of bones as well as muck, that were transported and produced instead in water or perhaps wind someplace.
8 0
2 years ago
Describe an experiment for the preparation and collection of Oxygen sodium peroxide​
vitfil [10]

Answer:

Making oxygen

Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:

hydrogen peroxide → water + oxygen

2H2O2(aq) → 2H2O(l) + O2(g)

The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.

Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.

To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.

Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.

5 0
3 years ago
A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var
RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:

<em>pV = nRT</em>

Solve for <em>p</em>: <em>p = nRT/V</em>

R = 8.314 kPa.L.K^(-1).mol^(-1)

<em>T</em> = (265 + 273.15) K = 538.15 K

<em>V</em> = 500.0 mL = 0.5000 L

∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

4 0
3 years ago
Read 2 more answers
In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t
Arada [10]

Answer:

        \large\boxed{\large\boxed{0.529M}}

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

       r=d[A]/dt=-k[A]

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

      [A]=[A]_0e^{-kt}

You know [A]₀, k, and t, thus you can calculate [A].

       [A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}

       [A]=0.529M

7 0
3 years ago
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