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2 years ago

Please help it’s due today I will give brainliest

Chemistry

1 answer:

Thepotemich [5.8K]2 years ago

7 0

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

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A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re

sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature 10.0°C + 273 = 283 K

T₁ = initial temperature , 40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0

3 years ago

What do we beileve is at the center of the milky way

BaLLatris [955]

Answer:

I think the answer is C because I know

5 0

2 years ago

Find the volume of 20g of H₂ at STP<br><br>

ANEK [815]

Answer:224

Explanation:

We should answer it with Stoichiometry

We say: 20 g H2× (1 mol/ 2g)× ( 22.4 lit/ 1 mol) = 224

Means: we have 20 grams and every 2g H2, equals to 1 mol of it and every 1 mol of H2, equals to 22.4 lit( because of STP)

hope you got this:)

6 0

2 years ago

The independent variable has control and affects the

Stella [2.4K]

Answer:

The independent variable is the condition that you change in an experiment. It is the variable you control.

Explanation:

It is called independent because its value does not depend on and is not affected by the state of any other variable in the experiment. Sometimes you may hear this variable called the "controlled variable" because it is the one that is changed.

3 0

3 years ago

Read 2 more answers

The graph above shows the changes in temperature recorded for the 2.00 l of h2o surrounding a constant-volume container in which

Otrada [13]

Answer:

25.2 kJ

Explanation:

The complete question is presented in the attached image to this answer.

Note that, the heat gained by the 2.00 L of water to raise its temperature from the initial value to its final value comes entirely from the combustion of the benzoic acid since there are no heat losses to the containing vessel or to the environment.

So, to obtained the heat released from the combustion of benzoic acid, we just calculate the heat required to raise the temperature of the water.

Q = mCΔT

To calculate the mass of water,

Density = (mass)/(volume)

Mass = Density × volume

Density = 1 g/mL

Volume = 2.00 L = 2000 mL

Mass = 1 × 2000 = 2000 g

C = specific heat capacity of water = 4.2 J/g.°C

ΔT = (final temperature) - (Initial temperature)

From the graph,

Final temperature of water = 25°C

Initial temperature of water = 22°C

ΔT = 25 - 22 = 3°C

Q = (2000×4.2×3) = 25,200 J = 25.2 kJ

Hope this Helps!!!

6 0

2 years ago