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Evgen [1.6K]
3 years ago
13

The graph below shows how solubility changes with temperature

Chemistry
2 answers:
Svetradugi [14.3K]3 years ago
5 0

Answer:

Option C) Na2SO4 and Na2HAsO4

Explanation:

The solubility of salts can be affected by many things (pH, temperature, etc), this is why when comparing the solubility of two or more salts it's important to establish the conditions.

In this case the graph was made for changes in Temperature, so the first thing is to identify the T=40°C and then, which curves cross at that point.

As can be seen, the <u>red curve</u> (Na2HAsO4) and the <u>blue curve</u> (Na2SO4 ) cross at 40°C indicating that those compounds have similar solubilities at that temperature.

torisob [31]3 years ago
4 0
By looking at the graph, you can determine the answer is C. Na2HAsO4 and Na2SO4.
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I think it is D I'm not positive

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Based on your answers to parts a, b, and c select the best lewis structure (electron-dot structure) for hoi.
vodomira [7]

Answer : The lewis dot structure includes the lone pair of electrons in any element and is helpful for defining the bond formation using the electrons.


In the molecule of HOI hydrogen is to the left of oxygen; oxygen is in middle and Iodine is at right of oxygen.


The picture is attached for better understanding.

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3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
Calculate the molarity of the sodium acetate solution as described below.
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Answer:

This question is incomplete.

Explanation:

This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;

Molarity = number of moles (n) / volume (in liter or dm³)

To calculate the number of moles of NaC₂H₃O₂, we say

number of moles (n) =

given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂

The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000

The unit for Molarity is M (Molar concentration), mol/L or mol/dm³

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What is the difference between a theory and a law?​
kondaur [170]

Answer:

theory is diffrent from law

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