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7nadin3 [17]
2 years ago
10

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

equally Among the 4 fifth grade classes in the school what fraction of the whole athletic field is reserved for each fifth class
Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
3 0

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

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The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the
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Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

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2 years ago
A liquid that occupies a volume of 4.7 liters has a mass of 5.1 kilograms what is the density of the liquid in kg/L
kotykmax [81]
Divide each by 4.7 to get ur kg/L which is 1.085kg/1L
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For nitrogen gas the values of Cv and Cp at 25°C are 20.8 J K−1 mol−1 and 29.1 J K−1 mol−1, respectively. When a sample of nitro
kifflom [539]

Answer:

The fraction of energy used to  increase the internal energy of the gas is 0.715

Explanation:

Step 1: Data given

Cv for nitrogen gas = 20.8 J/K*mol

Cp for nitrogen gas = 29.1 J/K*mol

Step 2:

At a constant volume, all the  heat will increase the internal energy of the gas.

At constant pressure, the gas expands and does work., if the volume changes.

Cp= Cv + R

⇒The value needed to change the internal energy is shown by Cv

⇒The work is given by Cp

To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp

Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol

Cv/Cp = 0.715

The fraction of energy used to  increase the internal energy of the gas is 0.715

3 0
2 years ago
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How many ethyne molecules are contained in 84.3 grams of ethyne (C2H2)?
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6 0
2 years ago
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a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin
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Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0
2 years ago
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