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2 years ago

What is the mass of 9.86×1024 molecules of co2? the molar mass of co2 is 44.01 g/mol.

1 answer:

7 0

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. And with the molar mass from moles to mass.

9.86×1024 molecules CO2 ( 1 mol / 6.022x10^23 molecules ) ( 44.01 g / 1 mol ) = 720.59 g CO2

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In Philip’s French class, the students are learning how to pronounce closed vowels and open vowels. The students are most likely

Sladkaya [172]

Answer:

It sounds like they are studying French phonemes

Explanations:

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2 years ago

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Water flowing at the rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54 to 430 kg/h of hot gas flo

weeeeeb [17]

Answer:

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

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2 years ago

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a

Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

Answer:

A. ∆H > 0, ∆S > 0, ∆G < 0

Explanation:

During boiling, a liquid is converted to vapour. This is a phase change for which heat is absorbed because energy must be taken in to break the intermolecular bonds in the liquid before it can be converted to a gas. Hence ∆H>0

Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

3 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc = 3.74*10⁻³

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = 3.74*10⁻³

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago

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