What is the equation for the base ionization constant of CH5P

Ann dissolves four tablespoons of salt in a bowl of water. She notices that the water has become salty. Ann then keeps

antoniya [11.8K]

Answer:

2)Both the salt dissolving and the evaporation of water are physical changes.

Explanation:

The two kinds of changes that has occurred are both physical changes.

Dissolution of a solute such as salt in water and evaporation of water from a solution are physical changes.

A physical change is one in which the physical property of matter is altered. Particularly the state and the form.

In many of these cases, the change is easily reversible.

Since this change is reversible in that we were able to collect the salt back, this is a physical change.

6 0

3 years ago

The amount of chemical weathering will increase if

olga2289 [7]

Chemical Weathering is the process of decomposition, erosion and spoilage of Solid materials like Building Material, Rocks e.t.c.
This decomposition takes place due to chemical reaction. The chemical present in Air along with water undergo chemical decomposition of Solid materials.
So, first requirement for this process is water, which is present in air called precipitation. Secondly, temperature also play important role.

Result:
<span>The amount of chemical weathering will increase if Temperature and Precipitation Increases.</span>

8 0

3 years ago

If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?

olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

HA H₃O⁺ A⁻

Initial, M 0.40 0 0

Change , M -x +x +x

Equilibrium, M 0.40 - x x x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0

3 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago

A boy lifts a 30N dragon 2 meters above the ground. How much work did the boy do on the dragon?

pychu [463]

Answer:

Work= Fs

Work 30x2

Work 60 J

Explanation:

A lot of work

3 0

3 years ago