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3 years ago

What is the energy change when the temperature of 14.8 grams of solid lead is decreased from 38.4 °C to 23.7 °C ?

Chemistry

1 answer:

zimovet [89]3 years ago

8 0

Answer: Q= -34.6 J

Explanation: Use Q= mc∆T

c for solid lead is equal to 0.159 J/g°C

Substitute the values:

=14.8 g(0.159J/g°C)(23.7°C-38.4 °C)

= 14.8 g(0.159J/g°C) ( -14.7°C)

= -34.6 J

Cancel out the following units: g and °C so the remaining unit is in J.

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Explain the laws of thermodynamics and why they matter.

creativ13 [48]

Following are the laws of thermodynamics, with suitable example.

1st Law of thermodynamics:
1st law of thermodynamics deals with conservation of energy. It is stated as ' energy cannot be created or destroyed in an isolated system'. According to this law, total energy of universe remains constant. Energy just gets converted from one form to another. For example, in case of burning of cracker, chemical energy stored in cracker is converted into heat, light and sound energy.

2nd Law of thermodynamics:
2nd law of thermodynamics deals with entropy change associated with system. It is stated as ' entropy of any isolated system always increases'. According to this law, the system tries to maximize entropy. System with higher entropy is more stable than system with lower entropy. For instance, at room temperature, ice melts into water, because water has higher entropy than ice. It may be noted that entropy is measure of disorder in system. Thus, higher the disorder in system, greater is the entropy.

3rd Law of thermodynamics:
3rd law of thermodynamics also deal with entropy change in system. According to this law, entropy of a system approaches a constant value as the temperature approaches absolute zero. This means that, as the temperature decreases, randomness in system decreases and finally at 0K, system is in highly order state, hence ideally system must have zero entropy. However, there is always some residual entropy present in system even at 0K, due structural orientation of molecules.

7 0

3 years ago

A sample of barium nitrate is placed into a jar containing water. The mass of the barium nitrate sample is 27g. Assume the water

34kurt

So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.

using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water

27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g

therefore,
you need 310.34g of water is in the jar.

5 0

4 years ago

Read 2 more answers

For many purposes we can treat butane (C4H10) as an ideal gas at temperatures above

masha68 [24]

Answer:

A. Yes

B. –176 °C

Explanation:

A. Yes

B. Determination of the new temperature of the gas.

Let the initial pressure be P

From the question given above, the following data were obtained.

Initial pressure (P1) = P

Initial temperature (T1) = 19 °C

Final pressure (P2) = ⅓ P1 = ⅓P = P/3

Final temperature (T2) =?

Next, we shall convert 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273

Initial temperature (T1) = 292 K

Since the volume is constant, we can obtain the new temperature of the gas as illustrated below:

Initial pressure (P1) = P

Initial temperature (T1) = 292 K

Final pressure (P2) = P/3

Final temperature (T2) =?

P1/T1 = P2/T2

P/292 = P/3 /T2

P/292 = P/3T2

Cross multiply

P × 3T2 = 292 × P

Divide both side by P

3T2 = (292 × P)/P

3T2 = 292

Divide both side by 3

T2 = 292/3

T2 = 97.33 ≈ 97 K

Finally, we shall convert 97 K to celcius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 97

T(°C) = 97 – 273

T(°C) = –176 °C

Thus, the new temperature of the gas is –176 °C.

6 0

3 years ago

Answer please hurry I need the answer

Nady [450]

Homologous chromosomes are chromosome with the same relative shape and gene location. However the allele composition may vary.

So I would say B, but only if the orange represents the same allele for a gene and not the same gene.

6 0

3 years ago

I will give you 100 points and mark you as brainilest if you answer the whole thing correctly.

Nookie1986 [14]

Answer:

1.) $Al_{2} O_{3}$

2.) $Li_{3} N\\$

3.) 2HCl

4.) MgO

5.) Mg and $O_{2}$

6.) $H_{2}$ and $O_{2}$

7.)Na and $Br_{2}$

8.) 3 aluminums and 3 chlorides

Explanation:

6 0

3 years ago