Which of the following is an example of a chemical property?

Need help what is the answer

WARRIOR [948]

Answer:

11.009306 or 11.009

Explanation:

boron 11 has atomic mass of 11.009306

6 0

3 years ago

The pressure of 8.40 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute

ollegr [7]

Answer:

33.6 L

Explanation:

From the question given above, the following data were obtained from:

Initial volume (V1) = 8.40 L L

Initial pressure (P1) = P

Initial temperature (T1) = T

Final pressure (P2) = one-half the original pressure = ½P

Final temperature (T2) = double the original temperature = 2T

Final volume (V2) =.?

Thus, the new volume of the gas can be obtained as follow:

P1V1/ T1 = P2V2 /T2

P × 8.40 / T = ½P × V2 /2T

P × 8.40 / T = P × V2 / 4T

Cross multiply

T × P × V2 = P × 8.40 × 4T

Divide both side by T × P

V2 = (P × 8.40 × 4T) /T × P

V2 = 8.40 × 4

V2 = 33.6 L

Thus, the new volume of the gas is 33.6 L

7 0

3 years ago

Part A

jeka94

The largest risks while designing a model to withstand a village include that the model does not mitigate the effects of the tsunami or only mitigates the effects partially, which would cause damages to the homes.

Designing a model to withstand the effect of any natural phenomenon such as an earthquake, fire or tsunami is not an easy task and will require the following cycle:

Designing the model.
Testing the model.
Making changes or designing a new model.

In the case of a model for tsunamis, it is likely the following problems occur:

The model does not protect the houses from tsunamis.
The model does not protect the houses completely.

This would lead to negative effects such as:

Damages in the houses.
Dead or injured people.
Destruction of infrastrcture.

Note: This question is incomplete because the context is missing; here is the missing part.

Protecting Your Model Village from Tsunamis this task, you will design a model village to withstand the effects of a tsunami.

Learn more about tsunami in: brainly.com/question/1126317

6 0

3 years ago

How many atoms of each element are in each element represented in 3 Fe(OH)2

Svetlanka [38]

Fe - 3
O - 6
H - 6
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7 0

4 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago