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Aleksandr-060686 [28]
3 years ago
5

2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vapori

zation of methanol at 64°C is 35.3 kJ mol-1. Find W, Q, U, and H for this process.
Chemistry
1 answer:
Sophie [7]3 years ago
4 0

Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

Explanation:

Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,  

Moles = Mass / Molar mass

= 64.0 grams / 32.0 grams per mole

= 2 mol

The amount of energy given by the process of condensation is,  

ΔH = 2 mol × 35.3 kJ/mol = 70.6 kJ

In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,  

W = -P ΔV

Let us first find the volume of 2.0 mole gas at 64 °C, or 64 + 273 = 337 K,  

PV = nRT

V = nRT/P

= 2 mol × 0.08206 L atm per mol K × 337 K/1 atm

= 55.3 L

As the liquid condenses in the process, the change in volume would be negligible. So, the volume change will be -55.3 L

W = - 1 atm × - 55.3 L

W = 55.3 L.atm

W = 55.3 L.atm × 101.3 J/1 L atm = 5602 J

W = 5602 × 1 kJ / 1000 J = 5.602 kJ

W = 5.602 kJ

Now U can be calculated using the formula,  

U = q + W

= -70.6 kJ + 5.602 kJ

= -65. kJ

Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.  

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Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

7 0
2 years ago
The Law of _____states that substances combine in predictable proportions and that excess reactants remain unchanged.
RideAnS [48]

Answer: definite proportions.


Explanation:


1) The definite proportions law states that compounds will always have the same kind of atoms (elements) in the same mass proportion (ratios).


2) For example, a molecule of water will alwys have the same mass ratio of hydrogen atoms to oxygen atoms. That is what permits to obtain the chemical formula of the water molecule as H₂O.


The mass of the two hydrogen atoms will be in a fixed ratio respect to the mass of the oxygen atoms.


Then, if you have one reactant in less proportion than the other, respect to the ratio stated by the chemical formula of water, the former will react completely (it is the limiting reactant) with the corresponding (proportional) mass of the later. Then there will be an excess of the later reactant which will not react (will remain unchanged).


The reactants can only react in the proportion defined by the chemical formulas of the final products.

4 0
3 years ago
Read 2 more answers
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
Which of the following statements describes sound energy?
Galina-37 [17]

Answer:

C.

Explanation:

8 0
2 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
3 years ago
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