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2 years ago

Compare bond lengths in butane and t butylcyclohexane

2 answers:

5 0

The bond angles of butane and tert butylcyclohexane are different. First, butane is an alkane that has four carbon atoms and its molecular geometry is that of a straight line. While tert butylcyclohexane is a cycloalkane that forms a ring like structure with one carbon atom attached with a tert butyl branch. Second, the tert butylcyclohexane is shorter in bond length and is strained due to the branched alkyl attached to the cycloalkyl compared to the butane that has no steric hindrance.

maks197457 [2]2 years ago

5 0

Answer:

The lengths of the C-C bonds increases with the decrease in the resistance of said bond, for example, a triple bond has a shorter length than in the case of a single bond. Butane has the single bonds and the CC bond is hybridized with sp3 hybridization, however in the butylcyclohexane structure the CC bond is also sp3 and the angle is 120°, however the angle shown is equal to 109.5°, so there is a certain angular tension and it is very unstable with respect to butane

Explanation:

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If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/

riadik2000 [5.3K]

Answer: The energy of the complex is $2.72\times 10^2kJ$

Explanation:

To calculate the energy of the complex, we use the equation given by Planck which is:

$\Delta E=\frac{N_Ahc}{\lambda}$

where,

$\lambda$ = Wavelength of the complex = $440nm=4.40\times 10^{-7}m$ (Conversion factor: $1m=10^9nm$ )

h = Planck's constant = $6.624\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

$\Delta E$ = energy of the complex

Putting values in above equation, we get:

$\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ$

Conversion factor used: 1 kJ = 1000 J

Hence, the energy of the complex is $2.72\times 10^2kJ$

5 0

2 years ago

Letter flipper on a game show. Periodic table pun

fomenos

Is there a question or is this a weird "Fun Fact"

5 0

2 years ago

Read 2 more answers

1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?

Lyrx [107]

Answer: Option (b) is the correct answer.

Explanation:

The energy necessary to remove an electron from a gaseous atom or ion is known as ionization energy.

This means that smaller is the size of an atom more amount of energy has to be supplied to it in order to remove the valence electron. This is because in small atom or element there will be strong force of attraction between the nucleus and electrons.

So, high amount of energy has to be supplied to remove the valence electrons.

As electronic configuration of helium is $1s^{2}$ . So, due to completely filled valence shell it is more stable in nature.

As a result, we need to provide very high amount of energy to remove an electron from a helium atom.

Thus, we can conclude that out of the given options helium element would the first ionization energy of the atom be higher than that of the diatomic molecule.

7 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0

2 years ago