Suppose that 3.33 g of acetone at 25.0 °C condenses on the surface of a 44.0-g block of aluminum that is initially at 25 °C. If the heat released during condensation goes only toward heating the metal, what is the final temperature (in °C) of the metal block?

Answers

Answer 1

Answer:

68.6 °C

Explanation:

From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'

Q = Q'

Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g

Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown

So, mL = m'c(θ₂ - θ₁)

θ₂ - θ₁ = mL/m'c

θ₂ = mL/m'c + θ₁

substituting the values of the variables into the equation, we have

θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C

θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C

θ₂ = 43.56 °C + 25 °C

θ₂ = 68.56 °C

θ₂ ≅ 68.6 °C

So, the final temperature (in °C) of the metal block is 68.6 °C.

Answer 2

The final temperature of the metal block is 74.97°C

What is the specific heat?

The specific heat of a substance is the required quantity of heat needed to raise the temperature of 1 gram of the substance by 1° C.

From the parameters given:

The mass of acetone = 3.33 g

The number of moles of acetone is:

[tex]\mathbf{= 3.33 g \times \dfrac{mol}{58.08 \ mol}}[/tex]

= 0.0573 mol

At standard conditions, the heat of vaporization of acetone is:

[tex]\mathbf{\Delta H = 32.0 \ kJ/mol \times 0.0578 \ mol } \\ \\ \mathbf{\Delta H = 1.8496 \ kJ } \\ \\ \mathbf{ \Delta H = 1.85 \times 10^3 \ J}[/tex]

Given that:

The mass of the metal (m) = 44.0 gThe initial temperature [tex]\mathbf{T_1}[/tex] = 25° CThe final temperature [tex]\mathbf{T_2 = ???}[/tex]

The specific heat of the aluminum is = 0.903 J/g° C

The heat energy can be computed as:

q = msΔT

q = 41 g × 0.903 J/g° C × (x - 25°C)

Using the calorimetry principle, heat energy lost by metal = heat energy gained by acetone.

i.e.

[tex]\mathbf{q_{(acetone)} gain = q_{(metal)} lost }[/tex]

[tex]\mathbf{-1.85 \times 10^3 \ J = - 41 g \times 0.903 \ J/g^0 C \times ( x - 25^0 c) }[/tex]

[tex]\mathbf{1.85 \times 10^3 \ J = 41 g \times 0.903 \ J/g^0 C \times ( x - 25^0 c) }[/tex]

[tex]\mathbf{(x - 25 ^0 C) = \dfrac{1.85 \times 10^3 \ J }{ 41 g \times 0.903 \ J/g^0 C}}[/tex]

[tex]\mathbf{(x - 25 ^0 C) = 49.97^0 C}}[/tex]

[tex]\mathbf{x = 49.97^0 C+25 ^0 C}}[/tex]

x = 74.97 °C

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Pls predict and balance the following Chemical equations 30points (5 per question)
Will mark Brainliest

1. Na2(SO4) (aq) + Ba(NO3)2(aq) -->

2. Na(NO3)(aq) + NH4Cl(aq) -->

4. ZnCl2 (aq) + K (SO4) (aq) -->

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1. Na2SO4 + Ba(NO3)2 → NaNO3 + BaSO4

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4. Sorry, I don't know

5. Refer to the attachment..

6. NaC2H3O2(aq) + AgNO3(aq) = NaNO3(aq) + AgC2H3O2

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[tex] \huge{\blue{Thank \: you}}[/tex]


1. Na2(SO4) (aq) + Ba(NO3)2(aq) -->
Answer: NaNO3 + BaSO4 - Chemical Equation Balancer.

2. Na(NO3)(aq) + NH4Cl(aq) -->
Answer: NaNO3 + NH4Cl → NaCl + NH4NO3 - Balanced equation

4. ZnCl2 (aq) + K (SO4) (aq) -->
Answer: (This is an impossible reaction)

5. ZnCl2 (aq) + Na2S (aq) -->
Answer: no reaction occurs

6. Na(C2H3O2) (aq) + Ag(NO3) (aq) -->
Answer: NaC2H3O2(aq) + AgNO3(aq) = NaNO3(aq) + AgC2H3O2

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Answer:

D. The nucleus occupies very little of the volume but contains most of its mass

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Answer:

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Answer:

Pb(OH)2 + 2 HCl ---> 2H2o + PbCl2

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I also have the same question

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Answers

Answer:

The object has to move a distance when a force is applied to it

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For work to be done on a body the force applied must move the body through a particular distance.

  Work done  = Force x distance  

If no distance is moved by the force, no work is done.

Also, the angle between the force and the distance must be 0 to do the maximum work on the body.

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Answers

Answer:

18.65004 grams H2O

Explanation:

First, we need to write down the balanced chemical equation for the decomposition reaction:

2LiOH -> H2O + Li2O

Since we have grams of LiOH and we need to know the grams of water, we need to convert to moles since we can only compare moles to moles.

The amu of LiOH is 23.947.

The given grams of LiOH is 63.. To convert to moles, we will divide 63 by 23.947..

This gives us 2.6310 moles LiOH..

To convert to moles of H2O (and later grams of H2O), we will use the mole fractions from the balanced equation...

When we look at the balanced equation we can see that 2 moles of LIOH can produce 1 mol of Water, so:

2.6310 moles [tex]* \frac{1 molH_{2}O}{2 mol LiOH}[/tex] = 1.3155 moles H2O

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According to stoichiometry of the chemical equation of decomposition of lithium hydroxide 23.6 g of water are produced from the decomposition of 63 grams of lithium hydroxide.

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Stoichiometry is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

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In the given example, as 47.9 g of lithium hydroxide produces 18 g water ,therefore 63 of lithium hydroxide will produce 63×18/47.9=23.6 g of water.

Thus, 23.6 g of water are produced from the decomposition of 63 grams of lithium hydroxide.

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how many grams is 5.16 mol br2?

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Answer:

Mass = 824.57 g

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Given data:

Mass of bromine = ?

Number of moles of bromine = 5.16 mol

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Mass = number of moles × molar mass

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