All categories

3 years ago

In the reaction between NaNO3 (aq) and Li2SO4 (aq), will a precipitate form?

Chemistry

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer: The correct answer is No, this reaction produces no insoluble products.

Explanation:

Precipitate is defined as the insoluble salt which emerges when two different solutions are mixed together. It settles down at the bottom of the flask after some time.

The balanced chemical equation for the reaction of sodium nitrate and lithium sulfate follows:

$2NaNO_3(aq.)+Li_2SO_4(aq.)\rightarrow Na_2SO_4(aq.)+2LiNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of aqueous solution of sodium nitrate reacts with 1 mole of aqueous solution of lithium sulfate to produce 1 mole of aqueous solution of sodium sulfate and 2 moles of aqueous solution of lithium nitrate.

All the products are aqueous solution and thus, no precipitate is formed.

You might be interested in

Why are Li Na and k Placed in the same group of the periodic table?

Rudiy27

Answer:

Their valence electrons/properties

Explanation:

Elements in the same group of the periodic table have the same amount of valence electrons, which determines how they bond with other elements in other groups. This also determines their properties such as their reactivity. Since Li, Na, and K are in the same group, they must have similar properties and the same amount of electrons in their outermost shell.

5 0

4 years ago

An atom has radius of 227 pm and crystallizes in a body-centered cubic unit cell. What is the volume of the unit cell

sdas [7]

We will see that the volume of the unit cell is 144,070,699.06 pm^3

<h3>How to get the volume of a body-centered cubic unit cell?</h3>

In a body-centered cubic unit cell, the side length of the cube is given as:

$S = \frac{4}{\sqrt{3} } *R$

Where R is the radius of the atom.

And the volume of a cube is the side length cubed, then we can see that the volume of our cube will be:

$V = S^3 = (\frac{4}{\sqrt{3} }*227pm)^3$

Solving that we get:

$V = (\frac{4}{\sqrt{3} }*227pm)^3 = 144,070,699.06pm^3$

This is the approximated volume of the unit cell.

If you want to learn more about unit cell structures, you can read:

brainly.com/question/13110055

5 0

2 years ago

Rationing is an attempt to limit shortages that naturally result from what?

makvit [3.9K]

Answer:

Price ceiling. In order to limit the shortages that naturally result from a price ceiling, the government can do what to certain goods? Ration.

Explanation:

5 0

3 years ago

What is the number of moles of solute in 250 mL of a 0.4 M solution ?

Levart [38]

.4 M means that in 1 Liter of solution there are .4 moles.

Molarity x Liters = Moles

We have molarity, so convert mL to liters

250 mL x 1000mL/1 Liter = .25 L

Now, set up the equation

.4 M x .25 L = .1 moles of solute

3 0

3 years ago

Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a

Alexandra [31]

Answer:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

Explanation:

1. Balanced chemical equation (given):

$2A(aq)+2B(aq)\rightleftharpoons 5C(aq)$

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

$K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

a. Increase [B]

Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant $K_{c}$ is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

b. Increase [A]

Rightward shift.

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

Leftward shift.

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for $K_{c}$ : by increasing the denominator the numerator has to increase to keep the same value of $K_{c}$

d. Decrease [A]

Leftward shift.

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

Leftward shift.

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

Rightward shift.

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

No shift

You need to perform some calculations and determine the reaction coefficient, $Q_c$ to compare with the equilibrium constant $K_{c}$ .

The expression for $Q_c$ has the same form of the equation for $K_{c}$ . but the it uses the inital concentrations instead of the equilibrium concentrations.

$Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

Doubling [A] and reducing [B] to one half would leave the product of [A]² by [B]² unchanged, thus $Q_c$ will be equal to $K_{c}$ .

When $Q_c$ = $K_{c}$ the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

Rightward shift.

Again, using the expression for $Q_c$ , you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that $Q_c$ will be greater than $K_{c}$ ..

When $Q_c$ > $K_{c}$ the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

Summarizing:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

4 0

4 years ago