The following aqueous solutions of are mixed: 100.0 mL of 1.00 M lithium bromide and 100.0 mL of 1.00 M lead (II) nitrate. In an organized and clear manner, show all of your work and answers for this problem on the uploaded work. (a) Write the molecular equation for this reaction. (b) Write the total ionic chemical equation for this reaction. (c) Write the net ionic chemical equation for this reaction. (d) Identify the spectator ions in this reaction. (e) Identify the limiting reactant. (f) Determine the mass of solid product that is formed?

Answer and Explanation:

(a) When lithium bromide (LiBr) solution is mixed with a solution of lead (II) nitrate (Pb(NO₃)₂), lithium nitrate (LiNO₃) and lead (II) bromide (PbBr₂) are formed, according to the following molecular equation:

2LiBr(aq) + Pb(NO₃)₂(aq) → 2LiNO₃(aq) + PbBr₂(s) ↓

As the product PbBr₂ is an insoluble solid, it precipitates (↓).

(b) The total ionic equation is written with all ions of the reaction - no matter if they participate in the precipitate formation or not:

2Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → 2Li⁺(aq) + 2NO₃⁻(aq) + PbBr₂(s)

(c) The net ionic equation is written including only the ions which participate in the precipitate formation. In this case, the precipitate is PbBr₂, and it is formed by Pb²⁺ and Br⁻ ions:

Pb²⁺(aq) + 2Br⁻(aq) → PbBr₂(s)

(d) The spectator ions are those which do not participate in the formation of the precipitate. From the total ionic equation, we can see that Li⁺ and NO₃⁻ ions are repeated on both sides of the equation, so they are redundant. Thus, the spectator ions are Li⁺ and NO₃⁻ ions.

(e) To identify the limiting reactant, we first calculate the moles of each compound as the product of the solution concentration and volume:

For LiBr:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of LiBr = 0.1 L x 1.00 mol/L = 0.1 mol

The same for Pb(NO₃)₂:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of Pb(NO₃)₂  = 0.1 L x 1.00 mol/L = 0.1 mol

From the total ionic equation, we can see that 2 mol of LiBr reacts with 1 mol of Pb(NO₃)₂ to give 1 mol of PbBr₂ (solid product). The stoichiometric molar ratio is 2 mol LiBr/1 mol Pb(NO₃)₂ and we have 0.1 mol of each reactant (0.1 mol LiBr/0.1 mol Pb(NO₃)₂= 1). As 2 mol LiBr/mol Pb(NO₃)₂ > 1 mol LiBr/mol Pb(NO₃)₂, LiBr is the limiting reactant.

(f) From the total ionic equation, we know that 2 moles of LiBr produce 1 mol of PbBr₂. To determine the mass of solid product (PbBr₂) formed, we first multiply the stoichiometric ratio (1 mol PbBr₂/2 mol LiBr) by the actual number of moles of limiting reactant we have (0.1 mol):

moles of PbBr₂ = 0.1 mol LiBr x (1 mol PbBr₂/2 mol LiBr) = 0.05 mol PbBr₂

Finally, we convert the moles of PbBr₂ to gram by using the molar mass of the compound:

Molar mass PbBr₂ = 207.2 g/mol + (2 x 79.9 g/mol) = 367 g/mol

grams of PbBr₂ = 0.05 mol x 367 g/mol = 18.35 g

Answer:

Step-by-step explanation:

The input force is 50 N. But it will not create not any change. No mechanical advantage is observed.

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.