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2 years ago

What atomic or hybrid orbitals make up the bond between c2 and o in acetaldehyde, ch3cho?

2 answers:

8 0

Acetaldehyde is an organic compound (a compound containing C atoms) composed of a carbonyl group. On the other hand, a carbonyl group is a functional group containing C = O. The hybrid orbitals of a compound determines the number pi and s orbitals in the electronic configuration. For a single bond, there are two s orbitals. For double bonds, on the other hand, the number of s orbital bond is 1 while the number of pi bonds is 2. For triple bonds, there are three pi bonds present in the cloud.

Thus for a c = O bond, the atomic orbital configuration is sp3 containing 1 s orbital and 2 pi bonds.

Margaret [11]2 years ago

7 0

Answer:

In terms of central atom carbon:

three sp2 hybridized orbitals

one "2p" orbital.

one sp2 hybridized orbital and one "2p" orbital bonds with oxygen atom.

Explanation:

In acetaldehyde we have two carbons.

One carbon is simple alkyl carbon

The other carbon is carbonyl carbon

The carbonyl carbon is bonded to the alkyl carbon and hydrogen by sigma bonds (one with each) and to the oxygen atom by one sigma and one pi bond.Which means that there are three sigma bonds and one pi bond.

The hybridisation of carbonyl carbon is sp2.

The atomic orbitals involved are:

three "2p" orbitals

one "2s" orbital

Out of the three "2p" orbitals : two are involved in hybridization.

The third one is showing side ways overlapping with the "2p" orbital of oxygen atom.

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At this point Ron is slightly confused, this isn’t surprising. However, Hermione is doing rather well with them. This also isn’t

Zigmanuir [339]

Answer:

$\boxed{\text{0.780 atm}}$

Explanation:

Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.

She shows Ron how to use the Ideal Gas Law to solve the problem.

pV = nRT

She collects the data:

V = 1.00 L; n = 0.0319 mol; T = 25.0 °C

She reminds him to convert the temperature to kelvins

T = (25.0 +273.15) K = 298.15 K

Then she shows him how to do the calculation.

$p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}$

Isn't she smart?

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2 years ago

Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K

DanielleElmas [232]

Answer:

Explanation:

The following equation relates to Born-Haber cycle

$\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L$

Where

$\triangle H_f$ is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

$IE_M$ is ionisation enthalpy of metal

$EA_X$ is electron affinity of non metal atom

$U_L$ is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

$U_L$ = - 707 KJ / mol

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2 years ago

A compound is known to contain only carbon, hydrogen, and oxygen. If the complete combustion of a 0.150-g sample of this compoun

Rzqust [24]

Answer: the empirical formula is C3H4O3

Explanation:Please see attachment for explanation

8 0

2 years ago

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1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )

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2 years ago

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The rate of disappearance of HCl was measured for the following reaction:

AURORKA [14]

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

after 430mins,t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

6 0

2 years ago