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Roman55 [17]
1 year ago
14

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

g condensation goes only toward heating the metal, what is the final temperature in celcius of he metal block? Specific heat capacity of aluminum is 0.903j/g C
Chemistry
1 answer:
Gwar [14]1 year ago
8 0

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.

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Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)

Explanation:

Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.

A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°

B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°

C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°

D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°

E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°

F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°

Hence Two bonding domains and two non bonding domains have the smallest bond angle.

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1 year ago
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1 year ago
1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?
larisa [96]

Answer:

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn

A)

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

B)

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

7 0
1 year ago
Read 2 more answers
A flask contains 1/3 mole of h2 and 2/3 mol of he. Compare the force on the wall per impact of h2 relative to that for he.
densk [106]

The force on the wall is actually the pressure exerted by gas molecules

Higher the pressure more the force exerted on the walls of container

The pressure depends upon the number of molecules of a gas

In a mixture of gas the pressure depends upon the mole fraction of the gas

As given the mole fraction of He is more than that of H2 therefore He will exert more pressure on the wall

The ratio of impact will be

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1 year ago
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The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a
shusha [124]

Answer:

See explanation

Explanation:

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specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

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More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%

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