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Vinil7 [7]
3 years ago
11

Help me please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
1 answer:
gulaghasi [49]3 years ago
5 0
Sunlight. Solar power!!
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Which is better disney plus or hulu​
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Answer:

disney plus

Explanation:

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3 years ago
What type of elements form ionic bonds with metals
patriot [66]

Answer:

between metal and nonmetal ions

Explanation:

8 0
2 years ago
Given the atomic weights of carbon, 12.01; hydrogen, 1.01; and oxygen, 16.0, what is the molar mass of glucose (C6H12O6)?
wel
Multiply them, then add total.

C 6 * 12.01 =

H 12 * 1.01 =

O 6 * 16.0 =
                  + ------------
molar mass =

Add them together to get molar mass
5 0
3 years ago
Read 2 more answers
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
all of the following electromagnetic waves possess high energy that will damage the human body if overexposed, except _____. ult
Marrrta [24]
We use radiowaves all the time because they are harmless.  (as far as we know, although bees are dying and know one knows why, perhaps all the waves we use do have a negative effect)
Can I get a brainliest, please?
3 0
3 years ago
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