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2 years ago

When the metal was placed in the calorimeter its

Chemistry

2 answers:

liq [111]2 years ago

7 0

Answer:

-79.6C

-80C

-81.2C

Explanation:

guajiro [1.7K]2 years ago

5 0

Answer:

Exam 3 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Explanation:

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Given that the vapor pressure of water is 17.54 torr at 20 °c, calculate the vapor-pressure lowering of aqueous solutions that a

Olenka [21]

1) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(sucrose - C₁₂O₂₂O₁₁) = b(solution) · m(H₂O).
n(C₁₂O₂₂O₁₁) = 1,8 mol/kg · 1 kg.
n(C₁₂O₂₂O₁₁) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol) = 0,968.
Raoult's Law: p(solution) = mole fraction of solvent · p(solvent).
p(solution) = 0,968 · 17,54 torr = 16,99 torr.
Δp = 17,54 torr - 16,87 torr = 0,55 torr.

2) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(NaCl) = b(solution) · m(H₂O).
n(NaCl) = 1,8 mol/kg · 1 kg.
n(NaCl) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
i(NaCl) = 2; Van 't Hoff factor. Because dissociate on one cation and one anions.

Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol · 2) = 0,94.

Raoult's Law: p(solution) = mole fraction of solvent · p(solvent)
p(solution) = 0,94 · 17,54 torr = 16,47 torr.
Δp = 17,54 torr - 16,47 torr = 1,06 torr.

7 0

3 years ago

Suppose you have two identical-looking metal spheres of the same size and the same mass. One of them is solid; the other is holl

stiv31 [10]

Answer:

We, apply equal torque to both spheres and measure their final angular acceleration.

Explanation:

The moment of Inertia of solid sphere will be (2/5)mr²

The moment of Inertia of hollow sphere will be (2/3)mr²

Thus, the moment of inertia of inertia of hollow sphere is greater than that of the solid sphere. Since, the torque is equal to the product of moment of inertia and angular acceleration.

Therefore, we design a test such that we apply the same amount of torque to both the spheres. Due to greater amount of moment of inertia the hollow sphere will acquire high angular acceleration, while solid sphere will acquire low angular acceleration.

Hence, by the measurement of final angular acceleration, we can determine, which sphere is which.

6 0

3 years ago

What is the activation energy of a reaction?

Svetradugi [14.3K]

Answer:

activation energy, in chemistry, the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport. In transition-state theory, the activation energy is the difference in energy content between atoms or molecules in an activated or transition-state configuration and the corresponding atoms and molecules in their initial configuration. The activation energy is usually represented by the symbol Ea in mathematical expressions for such quantities as the reaction rate constant, k = Aexp(−Ea/RT), and the diffusion coefficient, D = Doexp(−Ea/RT).

7 0

2 years ago

A mixture of CrBr3 and inert material is analyzed to determine the Cr content. First the mixture is dissolved in water. Then all

Masteriza [31]

Answer:

Mass% Cr = 85.5%

Explanation:

Given:

Mass of CrBr3 sample = 0.8409 g

Mass of the AgBr precipitate = 1.0638 g

To determine:

The mass percent of Cr in the sample

Calculation:

The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2

CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)

Molecular weight of AgBr =187.77 g/mol

Moles of AgBr precipitated is:

$Moles(AgBr)=\frac{Mass(AgBr)}{Mol.wt(AgBr)}=\frac{0.8409g}{187.77g/mol}=0.004478moles$

Since 1 mole of AgBr contains 1 mole of Cl, therefore:

# moles of Cl = 0.004478 moles

At wt of Cl = 35.45 g/mol

$Mass(Chloride)=moles*at.wt = .004478moles*34.45g/mol=0.1543$

$Mass%(chloride)=\frac{mass(chloride)}{mass(sample)}*100=\frac{0.1543}{1.0638}*100 = 14.50%$

Mass%(Cr) = 100 - 14.50=85.5%

5 0

3 years ago

The partial pressure of cartbon dioxide in the atmgsphere is 0239 toer Caiculate the partial pressure in mm He and atm Rpund eac

Verizon [17]

Answer :

The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

The pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

Explanation :

The conversion used for pressure from torr to mmHg is:

1 torr = 1 mmHg

The conversion used pressure from torr to atm is:

1 atm = 760 torr

or,

$1torr=\frac{1}{760}atm$

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.

Pressure in mmHg :

As, $1torr=1mmHg$

So, $0.239torr=\frac{0.239torr}{1torr}\times 1mmHg=0.239mmHg$

Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

Pressure in atm:

As, $1torr=\frac{1}{760}atm$

So, $0.239torr=\frac{0.239torr}{1torr}\times \frac{1}{760}atm=3.14\times 10^{-4}atm$

Thus, the pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

3 0

3 years ago