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meriva
3 years ago
15

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

er temperature than a solution containing 5.00 g of compound B in 100. g of water. Which compound has the greater molar mass
Chemistry
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

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enyata [817]

Answer:

NH4NO3 (s) --> NH4+ (aq) + NO3- (aq) H = +25.7 kJ/mol

Explanation:

Step 1: Data given

2NaHCO3(s) → Na2CO3 (s) + H2O (g) + CO2 (g) ΔH = –129 kJ

C6H6(s) → C6H6(l) ΔH=+9.87 kJ

NH4NO3(s )→ NH4+(aq) + NO3–(aq) ΔH=+25.7 kJ/mol

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) ΔH =–5,471 kJ/mol

Step 2:

2NaHCO3(s) → Na2CO3 (s) + H2O (g) + CO2 (g)  H = -129 kJ:

This is the decomposition reaction of NaHCO3.

NaHCO3 will break down and form Na2CO3, H2O and CO2

⇒ the heat here is heat of decomposition

C6H6 (s) --> C6H6 (l), H = +9.87 kJ  

In this reaction we'll not form a new product. Benzene will change phase. From its solid form to liquid benzene.

⇒ the heat here is heat of fusion

NH4NO3 (s) --> NH4+ (aq) + NO3- (aq) H = +25.7 kJ/mol  

When solid NH4NO3 is dossolved in water. it will dissociate in ammonium ion (NH4+) and nitrate ion (NO3-).

This reaction shows a solid dissolving in water and therefore the H represents the heat of solution.

⇒ The heat here is <u>heat of solution</u>

2C8H18 (l) + 25O2 (g) --> 16CO2 (g) + 18H2O (l), H = -5,471 kJ/mol: this is a combustion reaction

This is the combustion reaction of octane. CO2 and H2O will be produced.

⇒  The heat here is heat of combustion.

8 0
4 years ago
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

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Explanation:

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Answer:

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