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wariber [46]
3 years ago
6

A molecule has a carbon atom that is best described by sp2 hybridization. how many other atoms are connected to this carbon? thr

ee one four two
Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
<span>Three atoms can be connected to the carbon atom with sp2 hybridization of a molecule.As sp2 hybridization for three sp2 hybrid orbitals which can only be connected to three other atoms.As an example is the ethene molecule(C2H4) which have 2 Carbon atoms with sp2 hybridization and one Carbon atom is connected with two hydrogen atoms and one carbon atom.</span>
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This formula shows the chemical reaction that occurs when the body breaks down sugar. The products are carbon dioxide and water.
vodomira [7]
I would say that you would need 12 to have it be equal for this reaction to happen. I hope that this helped. :)
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3 years ago
Use the words protons neutrons and I suppose in the same sentence ​
PolarNik [594]

Answer:

Isotopes have the same number of protons but a different number of neutrons.

Explanation:

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3 years ago
Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed
masya89 [10]

(A) 7.28\cdot 10^{-10} m

The De Broglie wavelength of an electron is given by

\lambda=\frac{h}{p} (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

v=1.00\cdot 10^6 m/s

and its mass is

m=9.11\cdot 10^{-31} kg

So, its momentum is

p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s

And substituting into (1), we find its De Broglie wavelength

\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m

(B) 1.16\cdot 10^{-34}m

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s

And so, the De Broglie wavelength of the ball is given by

\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m

(C) 9.02\cdot 10^{-9}m

The location of the first intensity minima is given by

y=\frac{L\lambda}{a}

where in this case we have

y=0.492 cm = 4.92\cdot 10^{-3} m

L = 1.091 is the distance between the detector and the slit

a=2.00\mu m=2.00\cdot 10^{-6}m is the width of the slit

Solving the formula for \lambda, we find the wavelength of the electrons in the beam:

\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m

(D) 7.35\cdot 10^{-26}kg m/s

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

p=\frac{h}{\lambda}

where here we have

\lambda=9.02\cdot 10^{-9}m is the wavelength

Substituting into the formula, we find

p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s

7 0
3 years ago
Which profession is likely to use a radiation shield?
pantera1 [17]
<span>Which profession is likely to use a radiation shield?
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Read 2 more answers
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3 years ago
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