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earnstyle [38]
3 years ago
7

Determine whether you can swim in 1.00 x 10^27 molecules of water.​

Chemistry
1 answer:
zloy xaker [14]3 years ago
7 0

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

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nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?
storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

FeCl₃ dissolved in 450 mL of solution(will dissociate )

Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

  • mol FeCl₃(MW=162,2 g/mol)

\tt \dfrac{15}{162.2}=0.0925

  • mol Cl⁻ :

\tt \dfrac{3}{1}\times 0.0925=0.2775

  • molarity of Cl⁻ :

\tt \dfrac{0.2775}{0.45}=0.617~M

7 0
3 years ago
What is the primary cause of diffusion? 1.that some substances can dissolve in solvents 2.heat added to a substance 3.random int
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Diffusion : a process in which molecules intermingle as a result of  their kinetic energy of random motion
The primary cause of diffusion is : 3. Random internal motion of atoms and molecules

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6 0
3 years ago
Read 2 more answers
What is another word for chemical change
satela [25.4K]

oxidisation.

decomposition reaction.

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4 0
3 years ago
How much heat is lost when 1277 g of H20 at 400 K is changed into ice at 263 K? Plz help I’ll give you brainliest 20 points
OLga [1]

The heat lost is -7.32*10^-^5J

The heat lost when the ice is cooled from 400k to 263K can be calculated using the formula of heat transfer.

<h3>Heat Transfer</h3>

This is the heat transferred from a body of higher temperature to a body of lower temperature.

Q = mc(\delta)T

  • Q = Heat Transfer
  • m = mass = 1277g
  • ΔT  = change in temperature

\delta T = (400 - 273) - (263 - 273) = \\&#10;\delta T = T_2 - T_1\\&#10;\delta T = -10 - 127\\&#10;\delta T = -137^0C

We converted the temperature from kelvin scale into Celsius scale and find the change in temperature.

Solving for heat transfer

Q = mc\delta T\\&#10;Q = 1277 * 4.186 * -137\\&#10;Q = -732336.514J

The heat loss is approximately -7.32*10^-^5J

Learn more on heat transfer here;

brainly.com/question/16055406

6 0
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