How did altitude affect the freezing, melting, and boiling points of water?

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Chromatography is a pretty accurate description of what happens to ink on wet paper, because it literally means "color writing" (from the Greek words chroma and graphe). Really, though, it's a bit of a misnomer because it often doesn't involve color, paper, ink, or writing. Chromatography is actually a way of separating out a mixture of chemicals, which are in gas or liquid form, by letting them creep slowly past another substance, which is typically a liquid or solid. So, with the ink and paper trick for example, we have a liquid (the ink) dissolved in water or another solvent creeping over the surface of a solid (the paper).

The essential thing about chromatography is that we have some mixture in one state of matter (something like a gas or liquid) moving over the surface of something else in another state of matter (a liquid or solid) that stays where it is. The moving substance is called the mobile phase and the substance that stays put is the stationary phase. As the mobile phase moves, it separates out into its components on the stationary phase. We can then identify them one by one.

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

Step 2: Calculate the molar ratio

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

Final answer:

The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.

Explanation:

To solve this problem, we're going to use the atomic mass percentages to determine the empirical formula of the compound.

We do this by assuming we have a 100g sample of the compound. Therefore:

The mass of calcium (Ca) is 38.7g.

The mass of phosphorus (P) is 19.9g.

The mass of oxygen (O) is 41.2g.

Next, we calculate how many moles we have of each element:

• Ca: 38.7g / 40.08g/mol (the atomic mass of calcium) = 0.965 moles
• P: 19.9g / 30.97g/mol (the atomic mass of phosphorous) = 0.643 moles
• O: 41.2g / 16.00g/mol (the atomic mass of oxygen) = 2.575 moles

Then, we divide each of these numbers by the smallest number of moles, which is 0.643 (P):

• Ca: 0.965/0.643 = 1.5 (~1)
• P: 0.643/0.643 = 1
• O: 2.575/0.643 = 4

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Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

At the anode:

At oxidation reaction is:

The overall equation for the reaction is:

The overall cell potential is:

Using the formula for the Nernst equation:

where;

E = 0.66

(Zn^2+)=0.22 M

Then

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log  [H⁺]

4.118  = - 2  log  [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M =  0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 =  0.76 - 0.0592/2logQ

0.660 - 0.76  =  - 0.0592/2logQ

-0.1 =  - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 =  [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92  × 10^-3/[H^+]^2

[H^+]^2 = 7.92  × 10^-3/2.39 × 10^3

[H^+] = 1.82  × 10^-3

pH = -log[H^+]

pH = -log[ 1.82  × 10^-3]

pH = 2.7

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Answer:

1 litre of 1.0 M NaCl

Explanation:

When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;

NaCl(s) ------> Na^+(aq) + Cl^-(aq)

Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.

Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl

Final answer:

The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.

Explanation:

Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.

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Answer:

1. Mass of Carbon is 56.89g

2. Mass of Hydrogen is 6.33g

3. Mass of Oxygen is 75.88

Explanation:

The following were obtained from the question.

Mass of the compound = 139.1g

Mass of CO2 produced = 208.6g

Mass of H2O produced = 56.93

1. Determination of mass of Carbon (C). This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 208.6

Mass of C = 56.89g

2. Determination of the mass of Hydrogen (H). This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 56.93

Mass of H = 6.33g

3. Determination of the mass of oxygen (O).

This is illustrated below:

Mass of the compound = 139.1g

Mass of C = 56.89g

Mass of H = 6.33g

Mass of O = Mass of compound - (mass of C + Mass of H)

Mass of O = 139.1 - (56.89 + 6.33)

Mass of O = 139.1 - 63.22

Mass of O = 75.88

Answer:

Option d.

Explanation:

Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:

R-C(=O)-R'  

The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).  

The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.  

Hence, from all of the above we can discard the option a, (the carbonyl groups exhibits sp2 hybridization), the option b (carbon-oxygen bond is a bond polar) and the option c (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).

Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.

I hope it helps you!

Answer:

d. The functional group of this type of compound must always be in the middle of a carbon chain.

Explanation: