## How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g.

Question: How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g. Use correct number of significant digits;

Solution: Mol weight of Ca(NO3)2 = 90 + 28 +96
= 164 g/mol
100 g of Ca(NO3)2 = 0.6097 mol of Ca(NO3)2
1 mol of Ca(NO3)2 contains 6 number of oxygen atoms.
so, 0.6097 moles of Ca(NO3)2 contains = 0.6097 * 6 = 3.6582 Now, mole of each ‘O’ atom is 16 g/mol So, 100 g of Ca(NO3)2 contains = 3.6582*16 = 58.5312 g of oxygen.

## A superconductor is a substance that is able to conduct electricity without resistance

Question: A superconductor is a substance that is able to conduct electricity without resistance, a property that is very desirable in the construction of large electromagnets. Metals have this property if cooled to temperatures a few degrees above absolute zero, but this requires the use of expensive liquid helium (boiling point 4 K). Scientists have discovered materials that become superconductors at higher temperatures, but they are ceramics. Their brittle nature has so far prevented them from being made into long wires. A recently discovered compound of magnesium and boron, which consists of 52.9% Mg and 47.1% B, shows special promise as a high-temperature superconductor because it is inexpensive to make and can be fabricated into wire relatively easily. What is the formula of this compound?

Solution:

Let us take 100 g of compound
Then, mass of Mg = 52.9g
and mass of boron = 47.1g
Moles of Mg = Mass/Molar Mass = 52.9g/24.31g/mol
= 2.18 mol
and moles of B = 47.1g/10.81g/mol
= 4.36 mol
So, the molar ratio of Mg and B:
nMg/nB = 2.18 mol/4.36 mol = 1/2
Formula of the compound is MgB2

## Drag the “orbitals with electrons” box to the energy scale shown.

Question: Drag the “orbitals with electrons” box to the energy scale shown. Your goal in this exercise is to show nitrogen in an sp3-hybridized state which is capable of forming 3 bonds with three other atoms. Fill answer fields from left to right. Place orbitals with two electrons on the left (if any).

Solution: Concepts and reason

An orbital is the distribution of the probability density that is associated with the given electron centered at a given point. If the point is the nucleus of an atom, an atomic orbital is the resultant. If the point is the center of mass of multiple nuclei within a molecule, a molecular orbital is the resultant.

When an atom is required to bond with the other atoms to form a molecule, the valence atomic orbitals with almost similar energies ‘hybridize’ to form hybridized orbitals. These orbitals of identical shapes have identical associated energies.

The number of hybrid orbitals formed is equal to the number of atomic orbital hybridizing. Fundamentals

The hybridization of an atom in a molecule could be correlated with the number of bonds formed by it. This is tabulated as follows:

The sum of the principal quantum number and the azimuthal quantum number informs about the energy associated with a given orbital.

$(n+l)\left( {{\rm{n + l}}} \right)(n+l)$

Here

$n−{\rm{n}}\,{\rm{ - }}\,n$− Principal quantum number

$l−{\rm{l}}\,{\rm{ - }}l$− Azimuthal quantum number

The electronic configuration of nitrogen is $1s22s22p3{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}}1s22s22p3$.

The $(n+l)\left( {{\rm{n + l}}} \right)(n+l)$ values associated with the atomic orbitals involved are as follows:

$1s−12s−22p−3\begin{array}{l}\\{\rm{1s}}\,{\rm{ - }}\,{\rm{1}}\\\\{\rm{2s}}\,{\rm{ - }}\,2\\\\{\rm{2p}}\,{\rm{ - }}\,{\rm{3}}\\\end{array}1s−12s−22p−3​$

The atomic orbitals with lower $(n+l)\left( {{\rm{n + l}}} \right)(n+l)​$ values have lower associated energies. Therefore, they are filled first. In nitrogen, the orbital with the lowest associated energy, $1s{\rm{1s}}1s​$ , is filled foremost, followed by $2s{\rm{2s}}2s​$ , and finally ​$2p{\rm{2p}}2p ​$.

The hybridized orbitals possess a blend of characteristics associated with their constituent atomic orbitals. Therefore, they could be placed at an intermediate position between the respective atomic orbitals (here, $2sand2p{\rm{2s}}\,{\rm{and}}\,{\rm{2p}}2sand2p$ ).

Hence, the arrangement is given below:

The arrangement is given below:

The $sp3{\rm{s}}{{\rm{p}}^{\rm{3}}}sp3$ orbitals possess $25%{\rm{25\% }}25%$ ‘s’ character and $75%{\rm{75\% }}75%$ ‘p’ character. Correspondingly, the stabilization or the effective energy of the hybridized orbitals is a little lesser than the un-hybridized $′2p′{\rm{'2p'}}′2p′$ orbitals, but quite greater than the un-hybridized $′2s′{\rm{'2s'}}′2s′$orbitals. The electronic population of the un-hybridized, valence orbitals is 666 . Correspondingly, the hybridized orbitals also have the same population.

## A student attempts to identify an unknown compound by the method used in this experiment. She finds that when she heated a sample weighing 0.4862 g

1. A student attempts to identify an unknown compound by the method used in this experiment. She finds that when she heated a sample weighing 0.4862 g the mass barely changed, dropping to 0.4855 g. When the product was converted to a chloride, the mass went up, to 0.5247 g.

a. Is the sample a carbonate? NO

b. What are the two compounds that might be in the unknown? KHCO3 and NaHCO3

c. Write the chemical equation for the overall reaction that would occur when the original compound was converted to a chloride. If the compound is a hydrogen carbonate, use the sum of Reactions 1 and 2. If the sample is a carbonate, use Reaction 2. Write the equation for a sodium salt and then for a potassium salt.

d. How many moles of the chloride salt would be produced from one mole of original compound? _____

e. How many grams of the chloride salt would be produced from one molar mass of original compound? Molar masses: NaHCO3 84.008 g Na2CO3 105.99 g NaCl 58.44 g KHCO3 100.118 g K2CO3 138.21 g KCl 74.55 g

If a sodium salt, ____________ g original compound → ____________ g chloride I

f a potassium salt, ____________ g original compound → ____________ g chloride

f. What is the theoretical value of Q, if she has an Na salt? ____________ if she has a K salt? ____________

g. What was the observed value of Q? ____________

h. Which compound did she have as an unknown?

I have bolded my own answers. I need help with the rest, please show work!

Solution:

a) As decomposition temperature is very high ( barely change its weight at high temperature) then the compound is expected as carbonate .

So the first answer is “YES”

b) So, the probable compound is Na2CO3 or K2CO3  g)observed value is 0.5247g, close to expected value with K​​​​​​2​​​CO3​​ ,0.5345g.

h) So the compound was K​​​​​​2​​​​​CO​​​3  which also correlates with high decomposition temperature of this compound (1200°C).

## Student Study Guide for Chemistry (11th Edition): Chapter 2, Problem 81P

Question: What is wrong with the chemical formula for each of the following compounds: (a) magnesium iodate [Mg(I04)2], (b) phosphoric acid(H3P03),(c) barium sulfite (BaS), (d) ammonium bicarbonate (NH3HC03)?

Solution: (a) The formula should be Mg(IO3)2. IO3 is the iodate anion. IO4 is the periodate anion.

(b) Phosphoric acid is H3PO4, not H3PO3.

(c) Barium sulfite is BaSO3. SO3 is the sulfite anion.

(d) Ammonium bicarbonate is NH4HCO3. Ammonium is NH4, not NH3.

## separation and analysis procedure for the general unknown

A student performed the separation and analysis procedure for the general unknown (Group I, II, and III cations). Her initial separation results are summarized below: – treatment with HCl yielded a white precipitate (part 1) – treatment with thioacetamide yielded an orange/black precipitate (part 2) – a clear, blue/green solution (part 3) The white precipitate from part I was incubated in a hot water bath for 5 minutes. After incubation, the solid was gone. Treatment with potassium chromate resulted in a bright yellow/orange precipitate. The orange/black precipitate from part 2 was redissolved by acidification with nitric acid followed by treatment with ammonia. After the ammonia treatment, the solution turned dark blue. The blue/green solution from part 3 was treated with hydrogen peroxide and potassium hydroxide. A reddish precipitate formed. The mixture was centrifuged and the solid was separated from the solution. Treatment of the solid with ammonia had no effect. Acidification with hydrochloric acid followed by addition of potassium thiocyanate resulted in an intense red/orange solution. The solution from the initial centrifugation step to remove the hydroxide solid was clear and yellow. Treatment of this solution with barium chloride resulted in a yellow/white precipitate. Identify the ions present in the student’s mixture. Draw the appropriate flow chart for each cation (Group I, II, and III on the following pages. (page 2 and 3). Group I ion(s): Group II ion(s): Group III ion(s):

Solution: